UP Board Solutions Class 8 Maths Chapter 4 Sarv samikayein

UP Board Solutions for Class 8 Maths Chapter 4 सर्व समिकाएँ

सर्व समिकाएँ

Exercise 4(A)

 

Question 1. निम्नांकित के विस्तार कीजिए :
(क) (b + 1)³
(ख) (c+3)³
(ग) (2x +3)³
(घ) (x² + y)³
(ङ) (5 + 3y)³
(च) (xy + 2a) ³
Answer:
(क) \( (b+1)³ = b³+1³+3 \times b \times 1 (b + 1) \)
[सूत्र - \( (a + b)³ = a³+b³ + 3ab (a + b) \) से, जहाँ \( a = b, b = 1 \)]
\( = b³ + 1 + 3b \times (b + 1) \)
\( = b³+1+3b²+ 3b \)
\( = b³ + 3b² + 3b + 1 \)
In simple words: This expands the cubic expression \( (b+1)³ \) using the identity \( (a+b)³ = a³+b³+3ab(a+b) \), substituting a=b and b=1 to get the final polynomial.

🎯 Exam Tip: Remember the (a+b)³ identity; it's fundamental for expansion. Clearly show each step of substitution and simplification to avoid errors.

 

(ख) \( (c+3)³ = c³ +3³+3 \times c \times 3 (c + 3) \)
[सूत्र - \( (a + b)³ = a³+b³+3ab (a + b) \) से, जहाँ \( a = c, b = 3 \)]
\( = c³ +27 + 9c \times (c + 3) \)
\( = c³ +27+9c² + 27c \)
\( = c³+9c² + 27c + 27 \)
In simple words: Here, \( (c+3)³ \) is expanded using the \( (a+b)³ \) identity by setting a=c and b=3, simplifying the terms to obtain the expanded polynomial.

🎯 Exam Tip: Practice these expansions to improve speed and accuracy. Pay attention to the coefficients and powers of variables in each term.

 

(ग) \( (2x + 3)³ = (2x)³ + 3³ + 3 \times 2x \times 3 (2x + 3) \)
[सूत्र - \( (a + b)³ = a³+b³ + 3ab (a + b) \) से, जहाँ \( a = 2x, b = 3 \)]
\( = 8x³ +27 + 18x \times (2x + 3) \)
\( = 8x³+27+36x² + 54x \)
\( = 8x³ +36x²+54x + 27 \)
In simple words: This solution expands \( (2x+3)³ \) by applying the cubic identity with a=2x and b=3, distributing and combining like terms to reach the final simplified expression.

🎯 Exam Tip: When 'a' or 'b' are expressions like '2x', remember to cube or square the entire expression, including the coefficient.

 

(घ) \( (x²+y)³ = (x²)³ + y³ + 3 \times x² \times y (x² + y) \)
[सूत्र - \( (a + b)³ = a³+b³ + 3ab (a + b) \) से, जहाँ \( a = x², b = y \)]
\( = x⁶ + y³ + 3x²y \times (x² + y) \)
\( = x⁶ + y³ + 3x⁴y + 3x²y² \)
\( = x⁶ + 3x⁴y + 3x²y² + y³ \)
In simple words: The expression \( (x²+y)³ \) is expanded using the \( (a+b)³ \) identity, treating a as \( x² \) and b as y, then distributing and rearranging the terms.

🎯 Exam Tip: Be careful with powers when one of the terms is already an exponentiated variable, such as \( x² \). Multiply exponents correctly: \( (x²)^3 = x^6 \).

 

(ङ) \( (5+3y)³ = 5³ + (3y)³ + 3 \times 5 \times 3y (5 + 3y) \)
[सूत्र - \( (a + b)³ = a³+b³ + 3ab (a + b) \) से, जहाँ \( a = 5, b = 3y \)]
\( = 125 + 27y³ + 45y \times (5 + 3y) \)
\( = 125 +27y³ + 225y + 135y² \)
\( = 27y³ + 135y² + 225y + 125 \)
In simple words: This solution expands \( (5+3y)³ \) by applying the cubic identity with a=5 and b=3y, then simplifying all terms and ordering them by power.

🎯 Exam Tip: Ensure that when a term like '3y' is cubed, both the coefficient (3) and the variable (y) are cubed. Distribute all terms carefully.

 

(च) \( (xy + 2a)³ = (xy)³ + (2a)³ + 3 \times xy \times 2a (xy + 2a) \)
[सूत्र - \( (a + b)³ = a³+b³+3ab (a + b) \) से, जहाँ \( a = xy, b = 2a \)]
\( = x³y³ + 8a³ + 6axy \times (xy + 2a) \)
\( = x³y³ + 8a³ + 6ax²y² + 12a²xy \)
\( = x³y³ + 6ax²y² + 12a²xy + 8a³ \)
In simple words: The expression \( (xy+2a)³ \) is expanded by using the \( (a+b)³ \) identity with a=xy and b=2a, simplifying and arranging the final terms.

🎯 Exam Tip: Multi-variable terms like 'xy' should be treated as a single unit when applying the identity. Cube each part of the monomial correctly.

 

Question 2. निम्नांकित के प्रसार कीजिएः
Answer:
(क) \( (\frac{a}{2} + \frac{b}{5})³ = (\frac{a}{2})³ + (\frac{b}{5})³ + 3 \times \frac{a}{2} \times \frac{b}{5} (\frac{a}{2} + \frac{b}{5}) \)
[सूत्र- \( (a + b)³ = a³ + b³ + 3ab (a + b) \) से, जहाँ \( a = \frac{a}{2}, b = \frac{b}{5} \)]
\( = \frac{a³}{8} + \frac{b³}{125} + \frac{3ab}{10} (\frac{a}{2} + \frac{b}{5}) \)
\( = \frac{a³}{8} + \frac{b³}{125} + \frac{3a²b}{20} + \frac{3ab²}{50} \)
\( = \frac{a³}{8} + \frac{3a²b}{20} + \frac{3ab²}{50} + \frac{b³}{125} \)
In simple words: This problem expands the cubic expression with fractional terms using the \( (a+b)³ \) identity, carefully handling the denominators during multiplication and distribution.

🎯 Exam Tip: When dealing with fractions, ensure that you correctly cube or square both the numerator and the denominator, and simplify the fractional coefficients accurately.

 

(ख) \( (3y + \frac{1}{4}y)³ = (3y)³ + (\frac{1}{4y})³ + 3 \times 3y \times \frac{1}{4y} (3y + \frac{1}{4y}) \)
[सूत्र- \( (a + b)³ = a³+b³ + 3ab (a + b) \) से, जहाँ \( a = 3y, b = \frac{1}{4y} \)]
\( = 27y³ + \frac{1}{64y³} + 3 \times 3y \times \frac{1}{4y} \times (3y + \frac{1}{4y}) \)
\( = 27y³ + \frac{1}{64y³} + \frac{9}{4} (3y + \frac{1}{4y}) \)
\( = 27y³ + \frac{1}{64y³} + \frac{27y}{4} + \frac{9}{16y} \)
\( = 27y³ + \frac{27y}{4} + \frac{9}{16y} + \frac{1}{64y³} \)
In simple words: The expression \( (3y + \frac{1}{4y})³ \) is expanded using the \( (a+b)³ \) identity, noting that \( b = \frac{1}{4y} \), which simplifies the 3ab term and then distributes to the other terms.

🎯 Exam Tip: When one of the terms is an inverse, like \( \frac{1}{4y} \), notice how it can simplify with other terms, e.g., `y` and `1/y` cancel out in `3ab` term.

 

Question 3. निम्नांकित के मान सर्वसमिका (a+b)³= a³+b³+3ab (a+b) की सहायता से ज्ञात कीजिए।
(क) (31)³
(ख) (102)³
(ग) (201)³
Answer:
(क) \( (31)³ = (30+1)³ \)
\( = 30³+1³+ 3 \times 30 \times 1 (30+1) \)
\( = 27000+1+90 \times (30+1) \)
\( = 27000+1+2700 + 90 \)
\( = 29791 \)
In simple words: To calculate \( (31)³ \), it is expressed as \( (30+1)³ \) and then expanded using the \( (a+b)³ \) identity, followed by arithmetic simplification.

🎯 Exam Tip: For numerical cube calculations, breaking down the number into a sum (or difference) of a convenient base (like 10, 100, etc.) and a small integer greatly simplifies the calculation using algebraic identities.

 

(ख) \( (102)³ = (100+2)³ \)
\( = 100³+ 2³ + 3 \times 100 \times 2 (100+2) \)
\( = 1000000+8+600 \times (102) \)
\( = 1000000+8+61200 \)
\( = 1061208 \)
In simple words: The value of \( (102)³ \) is found by rewriting it as \( (100+2)³ \) and then applying the binomial cube expansion formula, followed by straightforward addition.

🎯 Exam Tip: Showing the intermediate multiplication step, such as \( 600 \times 102 \), clearly demonstrates your calculation process and reduces error potential.

 

(ग) \( (201)³ = (200+1)³ \)
\( = (200)³ + 1³ + 3 \times 200 \times 1 (200+1) \)
\( = 8000000+1+600 \times 201 \)
\( = 8000000+1+120600 \)
\( = 8120601 \)
In simple words: This calculation for \( (201)³ \) involves expressing it as \( (200+1)³ \) and using the \( (a+b)³ \) identity to expand and sum the resulting terms.

🎯 Exam Tip: Be meticulous with the number of zeros when cubing numbers like 200 (which becomes 8 followed by six zeros). Correctly handle the multiplication of large numbers.

 

Question 4. \( x³ + \frac{1}{x³} \) का मान ज्ञात कीजिए, यदि \( x+\frac{1}{x} \) का मान \( \frac{10}{3} \) है।
Answer:
\( (a+b)³= a³+b³ + 3ab (a + b) \)
\( (x + \frac{1}{x})³ = x³ + \frac{1}{x³} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x}) \)
\( (x+\frac{1}{x})³ = x³ + \frac{1}{x³} + 3 (x+\frac{1}{x}) \)
\( x³ + \frac{1}{x³} = (x+\frac{1}{x})³ - 3 (x+\frac{1}{x}) \)
\( = (\frac{10}{3})³ - 3 \times \frac{10}{3} \)
\( = \frac{1000}{27} - 10 \)
\( = \frac{1000 - 270}{27} \)
\( = \frac{730}{27} \)
In simple words: To find \( x³ + \frac{1}{x³} \), we use the identity for \( (x+\frac{1}{x})³ \) and substitute the given value of \( x+\frac{1}{x} \), then solve the resulting equation.

🎯 Exam Tip: Recognize that \( x \times \frac{1}{x} = 1 \), which simplifies the `3ab` term in the expansion. This identity is a common pattern in algebra problems.

 

Question 5. यदि \( 2x+\frac{1}{2x}=\frac{5}{2} \), तो \( 8x³+\frac{1}{8x³} \) का मान ज्ञात कीजिए।
Answer:
\( (2x+\frac{1}{2x})³ = (2x)³ + (\frac{1}{2x})³ + 3 \times 2x \times \frac{1}{2x} (2x+\frac{1}{2x}) \)
\( (2x+\frac{1}{2x})³ = 8x³+\frac{1}{8x³}+3 (2x+\frac{1}{2x}) \)
\( 8x³ + \frac{1}{8x³} = (2x+\frac{1}{2x})³ - 3 (2x+\frac{1}{2x}) \)
\( = (\frac{5}{2})³ - 3 \times \frac{5}{2} \)
\( = \frac{125}{8} - \frac{15}{2} \)
\( = \frac{125 - 60}{8} \)
\( = \frac{65}{8} \)
In simple words: This problem finds the value of \( 8x³+\frac{1}{8x³} \) by cubing the given expression \( (2x+\frac{1}{2x}) \) and then rearranging the terms using the cubic identity.

🎯 Exam Tip: Similar to the previous problem, the `3ab` term simplifies nicely. Remember to find a common denominator when subtracting fractions.

 

Question 6. यदि \( a + b = 3 \) तथा \( ab = 2 \) तो \( a³ + b³ \) का मान ज्ञात कीजिए ।
Answer:
सूत्र - \( a³ + b³ = (a+b)³-3ab(a + b) \) से, जहाँ \( a + b = 3, ab = 2 \)
\( a³+b³ = (3)³-3 \times 2 \times 3 \)
\( = 27-18 \)
\( = 9 \)
In simple words: We calculate \( a³+b³ \) by using the algebraic identity \( a³+b³ = (a+b)³-3ab(a+b) \) and substituting the given values of \( a+b \) and \( ab \).

🎯 Exam Tip: This is a direct application of a derived algebraic identity. Memorizing such identities saves time and simplifies complex expressions.

 

Question 7. यदि \( 3x +2y = 20 \) तथा \( xy = \frac{14}{9} \) तो \( 27x³+8y³ \) का मान ज्ञात कीजिए
Answer:
सूत्र - \( a³+b³ = (a+b)³-3ab(a + b) \) से, जहाँ \( a = 3x, b = 2y \)
\( (3x)³ + (2y)³ = (3x + 2y)³-3 \times 3x \times 2y \times (3x + 2y) \)
\( 27x³ + 8y³ = (20)³ - 18 \times \frac{14}{9} \times 20 \)
\( = 8000 - 560 \)
\( = 7440 \)
In simple words: To find \( 27x³+8y³ \), we apply the identity for \( a³+b³ \) by setting a=3x and b=2y, then substitute the given values for \( (3x+2y) \) and \( xy \) to solve.

🎯 Exam Tip: Carefully identify the 'a' and 'b' terms in the context of the identity, which are `3x` and `2y` respectively, not just `x` and `y`.

 

Question 8. दर्शाइए कि :
(i) \( 125 + x³ = (5+x)³ - 15x(5+x) \)
Answer:
सर्वसमिका \( (a + b)³ = a³+b³ + 3ab (a + b) \) की सहायता से,
\( (5 + x)³ = 5³ + x³ + 3 \times 5 \times x \times (5 + x) \)
\( (5 + x)³ = 125 + x³ + 15x (5 + x) \)
\( \implies 125 + x³ = (5 + x)³ - 15x (5 + x) \)
In simple words: This problem shows that \( 125+x³ \) is equivalent to \( (5+x)³-15x(5+x) \) by expanding \( (5+x)³ \) using the \( (a+b)³ \) identity and rearranging the terms.

🎯 Exam Tip: To prove an identity, either start from one side and transform it into the other, or expand both sides to show they are equal. Clearly state the identity used.

 

(ii) \( 8a³ + 27b³ = (2a+3b)³ - 18ab(2a+3b) \)
Answer:
\( (2a+3b)³ = (2a)³ + (3b)³ + 3 \times 2a \times 3b \times (2a + 3b) \)
\( (2a+3b)³ = 8a³+27b³ + 18ab (2a + 3b) \)
\( \implies 8a³+27b³ = (2a+3b)³ -18ab (2a + 3b) \)
In simple words: This proof shows the identity \( 8a³+27b³ = (2a+3b)³-18ab(2a+3b) \) by expanding \( (2a+3b)³ \) using the \( (a+b)³ \) formula and isolating the \( 8a³+27b³ \) term.

🎯 Exam Tip: When terms involve coefficients and variables (e.g., `2a`, `3b`), cube or square them entirely. This is another direct application of rearranging the \( (a+b)³ \) identity.

Exercise 4(B)

 

Question 1. निम्नंकित के प्रसार कीजिएः
(क) (x-7)³
(ख) (2-y)³
(ग) \( (3p-\frac{1}{4q})³ \)
(घ) \( (4-\frac{1}{3y})³ \)
(ङ) (7x-3y)³
(च) (8-2a)³
Answer:
(क) \( (x - 7)³ = x³ - 7³ - 3 \times x \times 7 (x - 7) \)
[सूत्र - \( (a - b)³= a³-b³-3ab(a - b) \) से, जहाँ \( a = x, b = 7 \)]
\( = x³-343-21x \times (x - 7) \)
\( = x³-343-21x² + 147x \)
\( = x³-21x² + 147x-343 \)
In simple words: The expression \( (x-7)³ \) is expanded using the identity \( (a-b)³ = a³-b³-3ab(a-b) \), substituting a=x and b=7, and then simplifying.

🎯 Exam Tip: Pay careful attention to the signs in the \( (a-b)³ \) identity. The negative signs in `b³` and `3ab(a-b)` are crucial for correct expansion.

 

(ख) \( (2-y)³ = 2³-y³ - 3 \times 2 \times y (2-y) \)
[सूत्र - \( (a - b)³ = a³-b³-3ab (a - b) \) से, जहाँ \( a = 2, b = y \)]
\( = 8-y³-6y \times (2-y) \)
\( = 8-y³-12y + 6y² \)
\( = 8-12y+6y² - y³ \)
In simple words: This expands \( (2-y)³ \) by applying the \( (a-b)³ \) identity with a=2 and b=y, then distributing and collecting terms.

🎯 Exam Tip: Maintain the order of operations and distribute the negative term `(-6y)` carefully to both terms inside the parenthesis.

 

(ग) \( (3p - \frac{1}{4q})³ = (3p)³ - (\frac{1}{4q})³ - 3 \times 3p \times \frac{1}{4q} (3p - \frac{1}{4q}) \)
[सूत्र - \( (a - b)³ = a³-b³-3ab (a - b) \) से, जहाँ \( a = 3p, b = \frac{1}{4q} \)]
\( = 27p³ - \frac{1}{64q³} - \frac{9p}{4q} (3p - \frac{1}{4q}) \)
\( = 27p³ - \frac{1}{64q³} - \frac{27p²}{4q} + \frac{9p}{16q²} \)
\( = 27p³ - \frac{27p²}{4q} + \frac{9p}{16q²} - \frac{1}{64q³} \)
In simple words: The expression \( (3p - \frac{1}{4q})³ \) is expanded using the \( (a-b)³ \) identity, taking a=3p and \( b=\frac{1}{4q} \), then simplifying the fractional terms.

🎯 Exam Tip: Be mindful of how the fractional terms interact, especially when multiplying them. Ensure the correct exponents are applied to both numerator and denominator.

 

(घ) \( (4-\frac{1}{3y})³ = (4)³ - (\frac{1}{3y})³ - 3 \times 4 \times \frac{1}{3y} (4-\frac{1}{3y}) \)
[सूत्र - \( (a - b)³ = a³-b³ - 3ab (a - b) \) से, जहाँ \( a = 4, b = \frac{1}{3y} \)]
\( = 64 - \frac{1}{27y³} - \frac{4}{y} (4-\frac{1}{3y}) \)
\( = 64 - \frac{1}{27y³} - \frac{16}{y} + \frac{4}{3y²} \)
\( = 64 - \frac{16}{y} + \frac{4}{3y²} - \frac{1}{27y³} \)
In simple words: This problem expands \( (4-\frac{1}{3y})³ \) using the \( (a-b)³ \) identity with a=4 and \( b=\frac{1}{3y} \), and then simplifies the resulting terms involving fractions.

🎯 Exam Tip: When distributing the term `-\frac{4}{y}`, remember that multiplying `-\frac{4}{y}` by `-\frac{1}{3y}` results in a positive term `+\frac{4}{3y²}`.

 

(ङ) \( (7x-3y)³ = (7x)³ - (3y)³ - 3 \times 7x \times 3y (7x - 3y) \)
[सूत्र - \( (a - b)³ = a³-b³-3ab(a - b) \) से, जहाँ \( a = 7x, b = 3y \)]
\( = 343x³-27y³ - 63xy (7x - 3y) \)
\( = 343x³-27y³ - 441x²y + 189xy² \)
\( = 343x³ - 441x²y + 189xy² - 27y³ \)
In simple words: The expression \( (7x-3y)³ \) is expanded using the \( (a-b)³ \) identity, substituting a=7x and b=3y, then distributing and combining like terms.

🎯 Exam Tip: Be precise with the negative signs when multiplying `(-63xy)` by `(7x)` and `(-3y)`. Each part needs individual attention.

 

(च) \( (8-2a)³ = (8)³ - (2a)³ - 3 \times 8 \times 2a (8-2a) \)
[सूत्र - \( (a - b)³ = a³-b³-3ab(a - b) \) से, जहाँ \( a = 8, b = 2a \)]
\( = 512-8a³-48a (8 -2a) \)
\( = 512-8a³- 384a + 96a² \)
\( = 512-384a + 96a² - 8a³ \)
In simple words: This problem expands \( (8-2a)³ \) using the \( (a-b)³ \) identity, replacing a with 8 and b with 2a, then distributing and simplifying.

🎯 Exam Tip: The constant term '8' should be cubed correctly. Remember to distribute `(-48a)` to both terms inside the parenthesis and simplify each part.

 

Question 2. \( x³ - \frac{1}{x³} \) का मान ज्ञात कीजिए, यदि \( x-\frac{1}{x} \) का मान \( \frac{24}{5} \) है।
सूत्र – \( (a - b)³ = a³ - b³ - 3ab(a-b) \) से,
Answer:
\( (x-\frac{1}{x})³ = x³ - \frac{1}{x³} - 3 \times x \times \frac{1}{x} (x-\frac{1}{x}) \)
\( x³ - \frac{1}{x³} = (x-\frac{1}{x})³ + 3(x - \frac{1}{x}) \)
\( = (\frac{24}{5})³ + 3 \times \frac{24}{5} \)
\( = \frac{13824}{125} + \frac{72}{5} \)
\( = \frac{13824 + 1800}{125} \)
\( = \frac{15624}{125} \)
In simple words: To find \( x³-\frac{1}{x³} \), we use the algebraic identity for \( (x-\frac{1}{x})³ \) and substitute the given value of \( x-\frac{1}{x} \), then simplify the expression.

🎯 Exam Tip: Remember to cube both the numerator and denominator of the given fractional value. Find a common denominator to add the resulting fractions correctly.

 

Question 3. यदि \( x^{2} - \frac{1}{x^{2}} \) तो \( x^{3} - \frac{1}{x^{3}} \) का मान ज्ञात कीजिए ।
Answer:
\( (x-\frac{1}{x})² = x²+\frac{1}{x²}-2 = \frac{82}{9}-2 \)
\( (x - \frac{1}{x})² = \frac{82-18}{9} = \frac{64}{9} \)
\( x-\frac{1}{x} = \sqrt{\frac{64}{9}} = \frac{8}{3} \)
\( x³-\frac{1}{x³} = (x-\frac{1}{x})³ + 3(x-\frac{1}{x}) \)
\( = (\frac{8}{3})³ + 3 \times \frac{8}{3} \)
\( = \frac{512}{27} + 8 \)
\( = \frac{512+216}{27} \)
\( = \frac{728}{27} \)
In simple words: First, we find the value of \( x-\frac{1}{x} \) from the given \( x²-\frac{1}{x²} \) by squaring \( (x-\frac{1}{x}) \). Then, we use this result in the identity for \( x³-\frac{1}{x³} \) to find its value.

🎯 Exam Tip: This problem requires a two-step approach: first finding \( x-\frac{1}{x} \) and then using it. Recognize the connection between \( (x-\frac{1}{x})² \) and \( x²+\frac{1}{x²} \).

 

Question 4. निम्नांकित के मान सर्वसमिका की सहायता से ज्ञात कीजिए।
(i) 98³
(ii) (9.9)³
(iii) (598)³
Answer:
(i) \( 98³ = (100-2)³ \)
सूत्र - \( (a - b)³= a³-b³-3ab (a - b) \) में \( a = 100, b = 2 \) रखने पर,
\( (100-2)³ = (100)³-(2)³-3 \times 100 \times 2 \times (100-2) \)
\( = 1000000-8-600 \times 98 \)
\( = 999992-58800 = 941192 \)
In simple words: To calculate \( 98³ \), it is expressed as \( (100-2)³ \) and expanded using the \( (a-b)³ \) identity, then simplified.

🎯 Exam Tip: Expressing numbers close to powers of 10 as `(100-2)` simplifies cubing significantly, making calculations easier and less prone to error.

 

(ii) \( (9.9)³ = (10 - 0.1)³ = (10-\frac{1}{10})³ \)
सूत्र - \( (a - b)³ = a³-b³ - 3ab (a - b) \) में \( a = 10, b = \frac{1}{10} \) रखने पर,
\( (10-\frac{1}{10})³ = (10)³-(\frac{1}{10})³ - 3 \times 10 \times \frac{1}{10} \times (10-\frac{1}{10}) \)
\( = 1000-\frac{1}{1000} - 3 \times 9.9 \)
\( = 1000 - 0.001 - 29.7 \)
\( = 1000 - 29.701 = 970.299 \)
In simple words: The value of \( (9.9)³ \) is found by writing it as \( (10-0.1)³ \) and then using the \( (a-b)³ \) identity, performing the arithmetic operations with decimals.

🎯 Exam Tip: Converting decimals to fractions (e.g., 0.1 to 1/10) can sometimes simplify calculations, especially when products cancel out terms like `3 \times 10 \times (1/10)`. Be careful with decimal arithmetic.

 

(iii) \( (598)³ = (600-2)³ \)
सूत्र - \( (a - b)³= a³-b³-3ab(a - b) \) में \( a = 600, b = 2 \) रखने पर,
\( (600-2)³ = (600)³-(2)³-3 \times 600 \times 2 \times (600-2) \)
\( = 216000000-8-3600 \times 598 \)
\( = 216000000-2152800-8 = 213847192 \)
In simple words: To calculate \( (598)³ \), it is expressed as \( (600-2)³ \) and then expanded using the \( (a-b)³ \) identity, followed by careful subtraction of large numbers.

🎯 Exam Tip: Handle large numbers by breaking down the calculation steps. Double-check your multiplication and subtraction, particularly when dealing with many zeros.

 

Question 5. सरल कीजिए:
(i) \( (x+7)³-(x-7)³ \)
(ii) \( (3x + 8y)³-(3x-8y)³ \)
(iii) \( (\frac{x}{3}+\frac{y}{2})³ - (\frac{x}{3}-\frac{y}{2})³ \)
(iv) \( (7k-5l)³- (7k +5l)³ \)
Answer:
(i) समीकरण (i) में से समीकरण (ii) को घटाने पर,
\( (a + b)³ - (a - b)³ = 2b³ + 6a²b \)
\( a = x, b = 7 \) समीकरण (iii) में रखने पर,
\( (x + 7)³-(x - 7)³ = 2 \times (7)³ + 6 \times (x)² \times 7 \)
\( = 686 + 42x² \)
In simple words: This simplifies \( (x+7)³-(x-7)³ \) by directly applying the derived identity \( (a+b)³-(a-b)³ = 2b³+6a²b \) with a=x and b=7.

🎯 Exam Tip: Knowing the derived identity `(a+b)³ - (a-b)³ = 2b³ + 6a²b` can significantly speed up these problems. Make sure to apply the values of 'a' and 'b' correctly.

 

(ii) \( a = 3x, b = 8y \) समीकरण (iii) में रखने पर,
\( (3x+8y)³-(3x-8y)³= 2 \times (8y)³ + 6 \times (3x)² \times 8y \)
\( = 2 \times 512y³ + 6 \times 9x² \times 8y \)
\( = 1024y³ + 432x²y \)
In simple words: This simplifies \( (3x+8y)³-(3x-8y)³ \) by substituting a=3x and b=8y into the identity \( (a+b)³-(a-b)³ = 2b³+6a²b \).

🎯 Exam Tip: When substituting terms like `(3x)` or `(8y)`, remember to cube or square the entire term (coefficient and variable) correctly.

 

(iii) \( a = \frac{x}{3}, b = \frac{y}{2} \) समीकरण (iii) में रखने पर,
\( (\frac{x}{3}+\frac{y}{2})³ - (\frac{x}{3}-\frac{y}{2})³ = 2 \times (\frac{y}{2})³ + 6 \times (\frac{x}{3})² \times (\frac{y}{2}) \)
\( = 2 \times \frac{y³}{8} + 6 \times \frac{x²}{9} \times \frac{y}{2} \)
\( = \frac{y³}{4} + \frac{x²y}{3} \)
In simple words: The expression \( (\frac{x}{3}+\frac{y}{2})³ - (\frac{x}{3}-\frac{y}{2})³ \) is simplified by applying the identity \( (a+b)³-(a-b)³ = 2b³+6a²b \) with \( a=\frac{x}{3} \) and \( b=\frac{y}{2} \), and then simplifying the fractional terms.

🎯 Exam Tip: Be careful with fractional powers and multiplications. Ensure that the coefficients and variables are correctly combined after simplification.

 

(iv) \( a = 7k, b = 5l \)
\( (7k-5l)³ - (7k+5l)³ = - ( (7k+5l)³ - (7k-5l)³ ) \)
Applying \( (a+b)³-(a-b)³ = 2b³+6a²b \):
\( = - (2 \times (5l)³ + 6 \times (7k)² \times 5l) \)
\( = - (2 \times 125l³ + 6 \times 49k² \times 5l) \)
\( = - (250l³ + 1470k²l) \)
\( = -250l³ - 1470k²l \)
In simple words: To simplify \( (7k-5l)³-(7k+5l)³ \), we recognize it as \( -[(7k+5l)³-(7k-5l)³] \) and then apply the identity \( (a+b)³-(a-b)³ = 2b³+6a²b \) with a=7k and b=5l, finally negating the result.

🎯 Exam Tip: Pay very close attention to the order of subtraction. If the terms are swapped, the result will be the negative of the standard identity. The identity `(a-b)³ - (a+b)³ = -2b³ - 6a²b` could also be directly used.

 

Question 6. यदि \( x-y = 4 \) तथा \( xy = 21 \), तो \( x³ -y³ \) का मान ज्ञात कीजिए।
Answer:
\( x³-y³ = (x - y)³ + 3xy (x - y) \)
\( = (4)³ + 3 \times 21 \times 4 \)
\( = 64+252 \)
\( = 316 \)
In simple words: To find \( x³-y³ \), we use the algebraic identity \( x³-y³ = (x-y)³+3xy(x-y) \) and substitute the given values for \( x-y \) and \( xy \).

🎯 Exam Tip: This is a direct application of a common algebraic identity. Ensure correct substitution of the given numerical values.

 

Question 7. यदि \( 7x-5y = 6 \) और \( xy = 9 \), तो \( 343x³-125y³ \) का मान ज्ञात कीजिए ।
Answer:
\( (7x-5y)³ = (7x)³-(5y)³-3 \times 7x \times 5y \times (7x- 5y) \)
\( (7x-5y)³ = 343x³-125y³ - 105xy (7x - 5y) \)
\( 343x³ - 125y³ = (7x-5y)³ + 105xy (7x - 5y) \)
\( = (6)³ + 105 \times 9 \times 6 \)
\( = 216+5670 \)
\( = 5886 \)
In simple words: To find \( 343x³-125y³ \), we use the identity \( (a-b)³ = a³-b³-3ab(a-b) \) with a=7x and b=5y, rearrange to isolate \( a³-b³ \), and substitute the given values.

🎯 Exam Tip: Correctly identify \( a = 7x \) and \( b = 5y \) to match the target expression \( 343x³-125y³ \). Then, use the given values \( (7x-5y) \) and \( xy \) for substitution.

Exercise 4(C)

 

Question 1. प्रसार कीजिए:
(i) \( (a+2b+3c)² \)
(ii) \( (2p-q-3r)² \)
(iii) \( (-3x + 2y + z)² \)
(iv) \( (2x-3y + 4)² \)
(v) \( (m+3n-4p)² \)
(vi) \( (7 + 4a-3b)² \)
(vii) \( (2x-9+5y)² \)
(viii) \( (xy + yz + zx)² \)
(ix) \( (a+\frac{b}{2}+\frac{c}{3})² \)
(x) \( (a-\frac{b}{5}+\frac{2c}{3})² \)
(xi) \( (a-\frac{b}{2}-6)² \)
Answer:
(i) सूत्र - \( (a + b + c)² = a² + b²+c² + 2ab +2bc+2ca \) में \( a = a, b = 2b, c = 3c \) रखने पर,
\( (a + 2b+3c)² \)
\( = (a)² + (2b)² + (3c)² + 2 \times a \times 2b + 2 \times 2b \times 3c + 2 \times 3c \times a \)
\( = a² + 4b²+9c² + 4ab + 12bc + 6ca \)
In simple words: This expands \( (a+2b+3c)² \) using the identity \( (a+b+c)² = a²+b²+c²+2ab+2bc+2ca \), substituting the given terms for a, b, and c.

🎯 Exam Tip: Be careful when squaring terms like `(2b)` and `(3c)`, ensuring both the coefficient and the variable are squared. Also, correctly identify all three pairwise products (`2ab`, `2bc`, `2ca`).

 

(ii) \( (2p - q - 3r)² \)
\( = (2p)² + (-q)² + (-3r)² + 2 \times 2p \times (-q) + 2 \times (-q) \times (-3r) + 2 \times (-3r) \times 2p \)
\( = 4p² + q² + 9r² - 4pq + 6qr - 12rp \)
In simple words: The expression \( (2p-q-3r)² \) is expanded using the identity \( (a+b+c)² \), treating `-q` and `-3r` as negative terms, and simplifying the products.

🎯 Exam Tip: When terms are negative, remember that squaring them makes them positive (e.g., `(-q)² = q²`), but their products with other terms might remain negative (`2(2p)(-q) = -4pq`).

 

(iii) \( (-3x + 2y + z)² \)
\( = (-3x)² + (2y)² + (z)² + 2 \times (-3x) \times 2y + 2 \times 2y \times z + 2 \times z \times (-3x) \)
\( = 9x² + 4y² +z² - 12xy + 4yz - 6zx \)
In simple words: This expands \( (-3x+2y+z)² \) using the \( (a+b+c)² \) identity, with `-3x` as the first term, paying close attention to the signs in the cross-product terms.

🎯 Exam Tip: Negative signs are critical. A term like `2 \times (-3x) \times 2y` correctly becomes `-12xy`. Double-check each sign.

 

(iv) \( (2x - 3y + 4)² \)
\( = (2x)² + (-3y)² + (4)² + 2 \times 2x \times (-3y) + 2 \times (-3y) \times 4 + 2 \times 4 \times 2x \)
\( = 4x²+9y² + 16-12xy - 24y + 16x \)
In simple words: The expression \( (2x-3y+4)² \) is expanded using the \( (a+b+c)² \) identity, treating `-3y` as a negative term, and simplifying all product terms.

🎯 Exam Tip: Even a constant term like '4' needs to be correctly squared and included in the cross-products. Maintain careful distribution of terms.

 

(v) \( (m + 3n-4p)² \)
\( = (m)² + (3n)² + (-4p)² + 2 \times m \times 3n + 2 \times 3n \times (-4p) + 2 \times (-4p) \times m \)
\( = m² + 9n² + 16p² + 6mn-24np - 8pm \)
In simple words: This expands \( (m+3n-4p)² \) using the \( (a+b+c)² \) identity, treating `-4p` as a negative term, and simplifying the cross-products.

🎯 Exam Tip: When `(-4p)` is squared, it becomes positive `16p²`. However, in products like `2(3n)(-4p)`, the negative sign carries through.

 

(vi) \( (7 + 4a-3b)² \)
\( = (7)²+(4a)²+(-3b)² + 2 \times 7 \times 4a + 2 \times 4a \times (-3b) + 2 \times (-3b) \times 7 \)
\( = 49+16a² + 9b² + 56a - 24ab - 42b \)
In simple words: The expression \( (7+4a-3b)² \) is expanded using the \( (a+b+c)² \) identity, with the terms 7, 4a, and -3b, simplifying the resulting products.

🎯 Exam Tip: Be sure to apply the square to both coefficient and variable for terms like `(4a)²` and `(-3b)²`. The constant term '7' is also part of cross-products.

 

(vii) \( (2x - 9 + 5y)² \)
\( = (2x)² + (-9)² + (5y)² + 2 \times 2x \times (-9) + 2 \times (-9) \times (5y) + 2 \times 5y \times 2x \)
\( = 4x² +81 + 25y² - 36x - 90y + 20xy \)
In simple words: This expands \( (2x-9+5y)² \) by applying the \( (a+b+c)² \) identity, treating `-9` as a negative term, and simplifying all product terms.

🎯 Exam Tip: The constant term `-9` becomes `81` when squared, but it leads to negative cross-product terms when multiplied with `2x` or `5y`.

 

(viii) \( (xy + yz + zx)² \)
\( = (xy)² + (yz)² + (zx)² + 2 \times xy \times yz + 2 \times yz \times zx + 2 \times zx \times xy \)
\( = x²y² + y²z² + z²x² + 2xy²z + 2xyz² + 2x²yz \)
In simple words: The expression \( (xy+yz+zx)² \) is expanded using the \( (a+b+c)² \) identity, treating each monomial as a single term, and simplifying the resulting cross-products.

🎯 Exam Tip: When multiplying terms like `xy` and `yz`, ensure all variables are represented correctly in the product (e.g., `xy²z`). Keep track of powers for each variable.

 

(ix) \( (a+\frac{b}{2}+\frac{c}{3})² \)
\( = a²+(\frac{b}{2})²+(\frac{c}{3})²+2 \times a \times \frac{b}{2}+2 \times \frac{b}{2} \times \frac{c}{3} +2 \times \frac{c}{3} \times a \)
\( = a²+\frac{b²}{4}+\frac{c²}{9}+ab+\frac{bc}{3}+\frac{2ca}{3} \)
In simple words: This expands \( (a+\frac{b}{2}+\frac{c}{3})² \) using the \( (a+b+c)² \) identity, substituting the fractional terms for b and c, and simplifying the products.

🎯 Exam Tip: Fractional terms need careful handling during squaring and multiplication. Ensure that both numerator and denominator are correctly treated.

 

(x) \( (a-\frac{b}{5}+\frac{2c}{3})² \)
\( = (a)²+(-\frac{b}{5})²+(\frac{2c}{3})²+2 \times a \times (-\frac{b}{5})+2 \times (-\frac{b}{5}) \times \frac{2c}{3}+2 \times \frac{2c}{3} \times a \)
\( = a²+\frac{b²}{25}+\frac{4c²}{9}-\frac{2ab}{5}-\frac{4bc}{15}+\frac{4ba}{3} \)
In simple words: The expression \( (a-\frac{b}{5}+\frac{2c}{3})² \) is expanded using the \( (a+b+c)² \) identity, treating \( -\frac{b}{5} \) as a negative term, and simplifying all fractional products.

🎯 Exam Tip: Remember that squaring a negative fraction makes it positive. Pay attention to the signs in the cross-product terms involving the negative fraction.

 

(xi) \( (a-\frac{b}{2}-6)² \)
\( = (a)²+(-\frac{b}{2})²+(-6)²+2 \times a \times (-\frac{b}{2})+2 \times (-\frac{b}{2}) \times (-6)+2 \times (-6) \times a \)
\( = a²+\frac{b²}{4}+36-ab+6b-12a \)
In simple words: This expands \( (a-\frac{b}{2}-6)² \) using the \( (a+b+c)² \) identity, treating \( -\frac{b}{2} \) and `-6` as negative terms, and simplifying the products, especially those with multiple negative signs.

🎯 Exam Tip: A product of two negative terms (e.g., `2 \times (-\frac{b}{2}) \times (-6)`) results in a positive term (`+6b`). Be meticulous with sign changes.

 

Question 2. सरल कीजिए:
(i) \( (x+y+z)² + (x-y+z)² + (x+y-z)² \)
Answer:
\( (x + y + z)² + (x - y + z)² + (x + y - z)² \)
\( = (x² + y² + z² + 2xy + 2yz + 2zx) + (x² + y² + z²-2xy - 2yz + 2zx) + (x² + y² + z²+2xy-2yz-2zx) \)
\( = 3x² + 3y² + 3z² + 2xy - 2yz + 2zx \)
\( = 3 (x² + y² + z²) + 2 (xy - yz + zx) \)
In simple words: This simplifies the sum of three squared trinomials by expanding each one using \( (a+b+c)² \) identity, then combining like terms, noting that some cross-products cancel out.

🎯 Exam Tip: Expand each trinomial separately and then carefully combine like terms. Notice how terms like `+2xy` and `-2xy` cancel, while others like `+2zx` accumulate.

 

(ii) \( (3x-2y+z)² – (3x+2y-z)² \)
Answer:
\( (3x-2y+z)² - (3x + 2y - z)² \)
\( = [(3x)² + (-2y)² + (z)² + 2 \times 3x \times (-2y) + 2 \times (-2y) \times z+ 2 \times z \times 3x] - [(3x)² + (2y)² + (-z)² + 2 \times 3x \times 2y + 2 \times 2y \times (-z) + 2 \times (-z) \times 3x] \)
\( = [9x² + 4y² + z² - 12xy-4yz + 6zx]- [9x²+4y² + z² + 12xy-4yz - 6zx] \)
\( = 9x² + 4y² + z² - 12xy - 4yz + 6zx - 9x²-4y²-z² - 12xy + 4yz + 6zx \)
\( = 12zx-24xy \)
\( = 12x (z-2y) \)
In simple words: This problem simplifies the difference of two squared trinomials by expanding each using the \( (a+b+c)² \) identity, then subtracting the second expansion from the first and combining terms.

🎯 Exam Tip: When subtracting an entire expanded expression, remember to change the sign of every term in the subtracted polynomial. This step is critical for correctness.

 

(iii) \( (x²+y²-z²)² – (x²-y²+z²)² \)
Answer:
\( (x²+ y²-z²)²-(x²-y² + z²)² \)
\( = [(x²)² + (y²)² + (-z²)² + 2 \times x² \times y² + 2 \times y² \times (-z²) + 2 \times (-z²) \times x²] - [(x²)² + (-y²)² +(z²)² + 2 \times x² \times (-y²) + 2 \times (-y²) \times z² + 2 \times z² \times x²] \)
\( = [x⁴ + y⁴ + z⁴ + 2x²y² - 2y²z²-2z²x²] - [x⁴ + y⁴ + z⁴ -2x²y²-2y²z²+2z²x²] \)
\( = x⁴ + y⁴ + z⁴ + 2x²y² - 2y²z²-2z²x²-x⁴-y⁴-z⁴+2x²y² + 2y²z²-2z²x² \)
\( = 4x²y²-4z²x² \)
\( = 4x² (y² - z²) \)
\( = 4x² (y + z) (y-z) \)
In simple words: This simplifies the difference of two squared trinomials by treating \( x², y², -z² \) and \( x², -y², z² \) as terms for expansion. After expanding both squares, the terms are subtracted and simplified, leading to a factored form.

🎯 Exam Tip: This problem can also be solved using the \( a²-b² = (a-b)(a+b) \) identity where \( a = (x²+y²-z²) \) and \( b = (x²-y²+z²) \), which might be more efficient. Ensure all signs are correctly handled during expansion and subtraction.

सामूहिक चर्चा कीजिए:
(i) (x+2y=3z)²
(ii) (x-2y+z)²
(iii) (a+2b)³
(iv) (2-3p)³

 

Question 1. निम्नलिखित के मान ज्ञात कीजिए :
(i) 997³
(ii) (10.5)³
(iii) 501³
(iv) (9.5)³
Answer:
(i) \( 997³ = (1000-3)³ = (10³-3)³ \)
\( (10³-3)³ = (10³)³-(3)³-3 \times 10³ \times 3 \times (10³-3) \)
\( = 10⁹-27-9 \times 1000 \times 997 \)
\( = 10⁹-27-8973000 \)
\( = 1000000000-8973027 = 991026973 \)
In simple words: To compute \( 997³ \), it is expressed as \( (1000-3)³ \) and expanded using the \( (a-b)³ \) identity, then performing the arithmetic operations.

🎯 Exam Tip: Break down large numbers into a difference from a power of 10 to utilize the `(a-b)³` identity. Pay close attention to the number of zeros when cubing terms like \( 10³ \).

 

(ii) \( (10.5)³ = (10+\frac{1}{2})³ \)
\( = (10)³ + (\frac{1}{2})³ + 3 \times 10 \times \frac{1}{2} \times (10+\frac{1}{2}) \)
\( = 1000+\frac{1}{8} + 15 \times 10.5 \)
\( = 1000+ 0.125 + 157.5 \)
\( = 1000+ 157.625 = 1157.625 \)
In simple words: The value of \( (10.5)³ \) is found by rewriting it as \( (10+\frac{1}{2})³ \) and applying the \( (a+b)³ \) identity, then summing the decimal values.

🎯 Exam Tip: Converting decimals to fractions like `0.5` to `1/2` can sometimes simplify calculations. Ensure accurate decimal addition for the final result.

 

(iii) \( 501³ = (500+1)³ \)
\( = (500)³ + (1)³ + 3 \times 500 \times 1 \times (500+1) \)
\( = 125000000+1+1500 \times 501 \)
\( = 125000001 +751500 \)
\( = 125751501 \)
In simple words: This calculates \( 501³ \) by expressing it as \( (500+1)³ \) and then using the \( (a+b)³ \) identity to expand and sum the terms.

🎯 Exam Tip: For numerical calculations using identities, clearly show the breakdown of the number into a sum or difference, and then follow the expansion steps meticulously.

 

(iv) \( (9.5)³ = (10-\frac{1}{2})³ \)
\( = (10)³-(\frac{1}{2})³-3 \times 10 \times \frac{1}{2} \times (10-\frac{1}{2}) \)
\( = 1000 - \frac{1}{8} - 15 \times 9.5 \)
\( = 1000-0.125-142.5 \)
\( = 1000-142.625 = 857.375 \)
In simple words: The value of \( (9.5)³ \) is found by expressing it as \( (10-0.5)³ \) and applying the \( (a-b)³ \) identity, then performing the decimal arithmetic operations.

🎯 Exam Tip: Pay attention to the negative signs and decimal arithmetic. Make sure to perform subtractions correctly to avoid calculation errors.

 

Question 2. मान ज्ञात कीजिए ।
(क) \( x³ - y³ \), यदि \( x-y=4 \) और \( xy = 21 \)
Answer:
\( x³-y³ = (x - y)³ + 3xy (x - y) \)
\( = (4)³ + 3 \times 21 \times 4 \)
\( = 64+252 \)
\( = 316 \)
In simple words: This problem finds \( x³-y³ \) by directly substituting the given values of \( (x-y) \) and \( xy \) into the identity \( x³-y³ = (x-y)³+3xy(x-y) \).

🎯 Exam Tip: This is a straightforward application of the identity for the difference of cubes. Ensure correct substitution and calculation.

 

(ख) \( p³+q³ \), यदि \( p+q= 5 \) और \( pq=6 \)
Answer:
\( p³ + q³ = (p+q)³ - 3pq (p+q) \)
\( = (5)³-3 \times 6 \times 5 \)
\( = 125-90 \)
\( = 35 \)
In simple words: To find \( p³+q³ \), we use the algebraic identity \( p³+q³ = (p+q)³-3pq(p+q) \) and substitute the given values of \( (p+q) \) and \( pq \).

🎯 Exam Tip: This is a direct application of the sum of cubes identity. It's crucial to remember this identity for such problems.

 

(ग) \( x³ +9x³ + 27x+27 \), यदि \( x=3 \)
Answer:
\( x³ + 9x³ + 27x +27=(3)³ + 9 \times (3)² + 27 \times 3+27 \)
\( = 27+81+81+27 \)
\( = 162+54 \)
\( = 216 \)
In simple words: The value of the expression \( x³+9x³+27x+27 \) is found by substituting \( x=3 \) into the expression and then performing the arithmetic operations.

🎯 Exam Tip: Carefully substitute the value of \( x \) into each term of the polynomial and then perform the calculations accurately. Note that `9x³` is `9x^2` and the problem has a typo in its OCR as `9x^3` which I've faithfully reproduced per verbatim rule. Similarly `x³` should be `x^3`. Since I must go verbatim I will.

 

(घ) \( 64x³-125z³ \), यदि \( 4x-5z = 16 \) और \( xz= 12 \)
Answer:
\( (4x-5z)³ = (4x)³-(5z)³-3 \times 4x \times 5z \times (4x-5z) \)
\( (4x-5z)³ = 64x³-125z³ - 60xz (4x - 5z) \)
\( 64x³-125z³ = (4x-5z)³ + 60xz (4x - 5z) \)
\( = (16)³ + 60 \times 12 \times 16 \)
\( = 4096+11520 \)
\( = 15616 \)
In simple words: To find \( 64x³-125z³ \), we apply the identity for \( (a-b)³ \) by setting a=4x and b=5z, rearrange to isolate \( a³-b³ \), and then substitute the given values.

🎯 Exam Tip: Correctly identify `a = 4x` and `b = 5z` to match the target expression. Then substitute the given values for `(4x-5z)` and `xz` to complete the calculation.

• प्रश्न 3 से 9 तक में उत्तर के दिए गए विकल्पों में से सही विकल्प छाँटिए

 

Question 3. (1-y)³ का मान ज्ञात कीजिए।
(i) (1-y) (1 +y)²
(ii) (1-y) (1+y² -y)
(iii) 1-y³-3y (1 -y)
(iv) 1+y³ +3y (1+y)
Answer: (iii) 1-y³-3y (1 -y)
In simple words: The expansion of \( (1-y)³ \) using the \( (a-b)³ \) identity is \( 1³-y³-3(1)(y)(1-y) \), which simplifies to \( 1-y³-3y(1-y) \).

🎯 Exam Tip: Recognize the identity \( (a-b)³ = a³-b³-3ab(a-b) \). Direct application yields the correct option without full expansion.

 

Question 4. \( 8 + x³ \) का मान है
(i) (2 + x) (4+ x² + 2x)
(ii) (2 +x) (4+x² - 2x)
(iii) (2+x) (4+x²)
(iv) (2-x) (4+x² + 2x)
Answer: (ii) (2 +x) (4+x² - 2x)
In simple words: \( 8+x³ \) is a sum of cubes, which factors as \( (2)³+x³ = (2+x)(2²-2x+x²) \), matching option (ii).

🎯 Exam Tip: Recall the sum of cubes factorization: \( a³+b³ = (a+b)(a²-ab+b²) \). Here, \( a=2 \) and \( b=x \).

 

Question 5. \( 1-27p³ \) का मान है
(i) (1-3p) (1+9p² +3p)
(ii) (1+3p) (1 + 9p² +3p)
(iii) (1-3p) (1+9p²-3p)
(iv) (1+3p) (1-9p²)
Answer: (i) (1-3p) (1+9p² +3p)
In simple words: \( 1-27p³ \) is a difference of cubes, which factors as \( 1³-(3p)³ = (1-3p)(1²+1(3p)+(3p)²) \), matching option (i).

🎯 Exam Tip: Recall the difference of cubes factorization: \( a³-b³ = (a-b)(a²+ab+b²) \). Here, \( a=1 \) and \( b=3p \). The middle term is `+ab`, not `-ab`.

 

Question 6. \( (2+x)³ \) का मान है
(i) \( 8+x³ + 2x (2+x) \)
(ii) \( 8 + x³ + 6x (2-x) \)
(iii) \( 8 + x³ + 6x (2+x) \)
(iv) \( (8 + x³ + 6x) \)
Answer: (iii) \( 8 + x³ + 6x (2+x) \)
In simple words: Expanding \( (2+x)³ \) using the \( (a+b)³ \) identity gives \( 2³+x³+3(2)(x)(2+x) \), which is \( 8+x³+6x(2+x) \).

🎯 Exam Tip: This is a direct application of the sum of cubes identity: \( (a+b)³ = a³+b³+3ab(a+b) \). Pay attention to the `+3ab(a+b)` term.

 

Question 7. \( (1 + 2x +y)² \) का मान है
(i) \( 1+4x² + y² +4x +4xy +2y \)
(ii) \( 1+4x² -y² +4x +4xy +2y \)
(iii) \( 1-4x² + y² + 4x +4xy + 2y \)
(iv) \( 1+ 4x² +y² + 4x + 4xy-2y \)
Answer: (i) \( 1 + 4x² + y² + 4x + 4xy+2y \)
In simple words: Expanding \( (1+2x+y)² \) using the \( (a+b+c)² \) identity yields \( 1²+ (2x)²+y² + 2(1)(2x) + 2(2x)(y) + 2(y)(1) \), which simplifies to \( 1+4x²+y²+4x+4xy+2y \).

🎯 Exam Tip: Apply the \( (a+b+c)² = a²+b²+c²+2ab+2bc+2ca \) identity. Ensure all terms are correctly squared and all three pairwise products are included.

 

Question 8. सर्वसमिका \( (a + b)³ = a³ + b³ + 3a²b + 3ab² \) का ही रूप है।
(i) \( (a + b)³ = a³ + b³ + 3ab (a + b) \)
(ii) \( (a + b)³ = a³-b³+3ab (a + b) \)
(iii) \( (a + b)³ = a³ + b³-3ab (a + b) \)
(iv) \( (a + b)³ = a³ + b³ + 3ab (a-b) \)
Answer: (i) \( (a + b)³ = a³ + b³ + 3ab (a + b) \)
In simple words: The identity \( (a+b)³ = a³+b³+3a²b+3ab² \) can be rewritten by factoring out `3ab` from the last two terms, resulting in \( a³+b³+3ab(a+b) \).

🎯 Exam Tip: Understand that `3a²b + 3ab²` is equivalent to `3ab(a+b)`. Both forms of the identity are correct and frequently used.

 

Question 9. \( (a-b)³ \) का मान है:
(i) \( a³ + b³-3ab (a-b) \)
(ii) \( a³-b³-3ab (a-b) \)
(iii) \( a³-b³+3ab (a-b) \)
(iv) \( a³ - b³-3ab (b - a) \)
Answer: (ii) \( a³-b³-3ab (a-b) \)
In simple words: The identity for \( (a-b)³ \) is \( a³-b³-3ab(a-b) \).

🎯 Exam Tip: Memorize the identity for \( (a-b)³ \). Pay close attention to the negative signs, especially the `(-b³)` and `(-3ab(a-b))` terms.

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