UP Board Solutions Class 9 Maths Chapter 4 Algebraic Identities Ex 4.4

Exercise 4.4 Algebraic Identities अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Questions)

Question 1. यदि a + b = 10 व ab = 21, तब a3 + b3 का मान ज्ञात कीजिए ।
Answer:
हलः
\(a + b = 10\)
घन करने पर
\(a^3 + b^3 + 3ab (a + b) = 1000\)
\(a^3 + b^3 + 3 \times 21(10) = 1000\)
\(\therefore a^3 + b^3 = 1000 - 630 = 370\)
In simple words: This problem uses the algebraic identity \((a+b)^3 = a^3+b^3+3ab(a+b)\). We substitute the given values of \(a+b\) and \(ab\) into this identity to find the value of \(a^3+b^3\).

🎯 Exam Tip: Remember to correctly apply the cubic identity for a sum and perform careful arithmetic calculations to avoid errors.

Question 2. यदि a + b = 8 व ab = 6, तब a3 + b3 का मान ज्ञात कीजिए।
Answer:
हल: ::
\(a + b = 8 \quad \quad \quad \quad \quad \quad \quad \quad (1)\)
घन करने पर
\(a^3 + b^3 + 3ab (a + b) = 512\)
\(a^3 + b^3 + 3 \times 6(8) = 512\)
\(a^3 + b^3 = 512 - 144 = 368\)
In simple words: Similar to the previous problem, this question also uses the identity \((a+b)^3 = a^3+b^3+3ab(a+b)\). By plugging in the provided values for \(a+b\) and \(ab\), we can directly calculate \(a^3+b^3\).

🎯 Exam Tip: Practice recognizing when to apply the sum of cubes identity. Accuracy in substitution and basic arithmetic is crucial for full marks.

Exercise 4.4 Algebraic Identities लघु उत्तरीय प्रश्न (Short Answer Type Questions)

Question 3. यदि x + y = 10 व xy = 16, तब x2 - xy + y² व x2 + xy + y² के मान ज्ञात कीजिए।
Answer:
हलः
\(x^2 - xy + y^2 = x^2 + y^2 + 2xy - 3xy\)
\(= (x + y)^2 - 3xy = (10)^2 - 48 = 100 - 48 = 52\)
\(x^2 + xy + y^2 = (x + y)^2 - xy\)
\(= (10)^2 - 16 = 100 - 16 = 84\)
In simple words: This problem utilizes the identity \((x+y)^2 = x^2+y^2+2xy\). We rearrange the given expressions to use this identity, then substitute the values of \(x+y\) and \(xy\) to find their solutions.

🎯 Exam Tip: Knowing how to manipulate algebraic expressions like \(x^2+y^2 = (x+y)^2-2xy\) or \(x^2+y^2 = (x-y)^2+2xy\) is key for such problems.

Question 4. यदि x - y = 6 व xy = 20, तब x3 - y³ के मान ज्ञात कीजिए ।
Answer:
हलः
\(\therefore x^2 + y^2 + xy = (x - y)^2 + 3xy\)
\(= (6)^2 + 3 \times 20 = 36 + 60 = 96\)
\(\therefore x^3 - y^3 = (x - y)(x^2 + y^2 + xy) = (6)(96) = 576\)
In simple words: This problem involves two main algebraic identities: \((x-y)^2 = x^2+y^2-2xy\) to find \(x^2+y^2\), and the difference of cubes identity \((x^3-y^3) = (x-y)(x^2+xy+y^2)\). By calculating \(x^2+y^2+xy\) first, we can then find \(x^3-y^3\).

🎯 Exam Tip: When evaluating \(x^3-y^3\), remember to first find \(x^2+xy+y^2\). Often, the expression \(x^2+y^2+xy\) can be written in terms of \((x-y)^2\) and \(xy\).

Question 5. निम्न गुणनफल ज्ञात कीजिए।
(i) \( (3/x + 5/y)(9/x^2 + 25/y^2 - 15/xy) \)
(ii) \( (1-x)(1+x+x^2) \)
(iii) \( (x^2 - 1)(1 + x^2 + x^4) \)
(iv) \( (7m^4 + q)(49m^8 - 7m^4q + q^2) \)
(v) \( (1+y)(1-y + y^2) \)
Answer:
हलः
(i) \( (3/x + 5/y)(9/x^2 + 25/y^2 - 15/xy) = (3/x)^3 + (5/y)^3 = 27/x^3 + 125/y^3 \)
(ii) \( (1-x)(1+x+x^2) = 1-x^3 \)
(iii) \( (x^2-1)(1+x^2+x^4) = (x^2-1)((x^2)^2 + x^2 \cdot 1 + 1^2) = (x^2)^3 - 1^3 = x^6 - 1 \)
(iv) \( (7m^4 + q)(49m^8 - 7m^4q + q^2) = (7m^4)^3 + (q)^3 = 343m^{12} + q^3 \)
(v) \( (1+y)(1-y+y^2) = 1 + y^3 \)
In simple words: This question requires applying various algebraic identities for multiplication, such as sum/difference of cubes (\(a^3 \pm b^3\)) and sum/difference of powers. Each part simplifies using a specific identity.

🎯 Exam Tip: Familiarize yourself with common algebraic identities like \(a^3+b^3 = (a+b)(a^2-ab+b^2)\) and \(a^3-b^3 = (a-b)(a^2+ab+b^2)\) as they are frequently tested.

Question 6. सर्वसमिकाओं का प्रयोग करके निम्न गुणनफलों के मान ज्ञात कीजिए । (दिया है: x = 3 व y = -1)
(i) \( (x/3 + 5x^2)(25/x^2 - 25 + 25x^2) \)
(ii) \( (3/x - x/3)(9/x^2 + x^2/9 + 1) \)
(iii) \( (x/4 + y/3)(x^2/16 + y^2/9 - xy/12) \)
Answer:
हलः
(i) \( (x/3 + 5x^2)(25/x^2 - 25 + 25x^2) = (x/3)^3 + (5x^2)^3 \)
\(= x^3/27 + 125x^6 \)
\(x = 3\) रखने पर
\(= (3)^3/27 + (5 \times 3^2)^3 \)
\(= 27/27 + (5 \times 9)^3 \)
\(= 1 + (45)^3 = 1 + 91125 = 91126 \)
(ii) \( (3/x - x/3)(9/x^2 + x^2/9 + 1) = (3/x)^3 - (x/3)^3 \)
\(= 27/x^3 - x^3/27 \)
\(x = 3\) रखने पर
\(= 27/3^3 - 3^3/27 \)
\(= 27/27 - 27/27 = 1 - 1 = 0 \)
(iii) \( (x/4 + y/3)(x^2/16 + y^2/9 - xy/12) = (x/4)^3 + (y/3)^3 \)
\(= x^3/64 + y^3/27 \)
\(x = 3\) व \(y = -1\) रखने पर
\(= (3)^3/64 + (-1)^3/27 \)
\(= 27/64 - 1/27 \)
\(= (27 \times 27 - 1 \times 64) / (64 \times 27) = (729 - 64) / 1728 = 665 / 1728 \)
In simple words: This question involves applying the sum or difference of cubes identities and then substituting given values of \(x\) and \(y\) to find the final numerical answer for each expression.

🎯 Exam Tip: Always evaluate the expression using identities first, then substitute the values of variables to simplify calculations and reduce potential errors.