Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.3

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.3

Question 1. Expand
(i) \( (3m + 5)² \)
(ii) \( (5p - 1)² \)
(iii) \( (2n - 1)(2n + 3) \)
(iv) \( 4p² - 25q² \)
Answer:
(i) To expand \( (3m + 5)² \), we use the algebraic identity \( (a + b)² = a² + 2ab + b² \). Here, \( a = 3m \) and \( b = 5 \).
\( (3m + 5)² = (3m)² + 2(3m)(5) + 5² \)
\( = 3² m² + 30m + 25 \)
\( = 9m² + 30m + 25 \)
(ii) To expand \( (5p - 1)² \), we use the algebraic identity \( (a - b)² = a² - 2ab + b² \). Here, \( a = 5p \) and \( b = 1 \).
\( (5p - 1)² = (5p)² - 2(5p)(1) + 1² \)
\( = 5²p² - 10p + 1 \)
\( = 25p² - 10p + 1 \)
(iii) To expand \( (2n - 1)(2n + 3) \), we use the algebraic identity \( (x + a)(x + b) = x² + (a + b)x + ab \). Here, \( x = 2n \), \( a = -1 \) and \( b = 3 \).
\( (2n - 1)(2n + 3) = (2n)² + (-1 + 3)(2n) + (-1)(3) \)
\( = 2² n² + 2(2n) - 3 \)
\( = 4n² + 4n - 3 \)
(iv) To expand \( 4p² - 25q² \), we recognize it as a difference of squares. We can rewrite it as \( (2p)² - (5q)² \), which fits the identity \( a² - b² = (a + b)(a - b) \). Here, \( a = 2p \) and \( b = 5q \).
\( 4p² - 25q² = (2p)² - (5q)² \)
\( = (2p + 5q)(2p - 5q) \)
In simple words: We used different math rules, called identities, to open up each expression. These rules help us multiply the terms quickly.

🎯 Exam Tip: Always identify the correct algebraic identity (like \( (a+b)² \), \( (a-b)² \), \( (x+a)(x+b) \) or \( a²-b² \)) before starting to expand; this is crucial for accurate results.

Question 2. Expand
(i) \( (3 + m)³ \)
(ii) \( (2a + 5)³ \)
(iii) \( (3p + 4q)³ \)
(iv) \( (52)³ \)
(v) \( (104)³ \)
Answer:
(i) To expand \( (3 + m)³ \), we use the identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = 3 \) and \( b = m \).
\( (3 + m)³ = 3³ + 3(3)²(m) + 3(3)m² + m³ \)
\( = 27 + 27m + 9m² + m³ \)
\( = m³ + 9m² + 27m + 27 \)
(ii) To expand \( (2a + 5)³ \), we use the identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = 2a \) and \( b = 5 \).
\( (2a + 5)³ = (2a)³ + 3(2a)²(5) + 3(2a)(5)² + 5³ \)
\( = 2³a³ + 3(2²a²)(5) + 6a(25) + 125 \)
\( = 8a³ + 60a² + 150a + 125 \)
(iii) To expand \( (3p + 4q)³ \), we use the identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = 3p \) and \( b = 4q \).
\( (3p + 4q)³ = (3p)³ + 3(3p)²(4q) + 3(3p)(4q)² + (4q)³ \)
\( = 3³p³ + 3(9p²)(4q) + 9p(16q²) + 4³q³ \)
\( = 27p³ + 108p²q + 144pq² + 64q³ \)
(iv) To find \( (52)³ \), we can write \( 52 \) as \( (50 + 2) \). Then we use the identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = 50 \) and \( b = 2 \).
\( (50 + 2)³ = 50³ + 3(50)²(2) + 3(50)(2)² + 2³ \)
\( = 125000 + 6(2500) + 150(4) + 8 \)
\( = 125000 + 15000 + 600 + 8 \)
\( = 140608 \)
(v) To find \( (104)³ \), we write \( 104 \) as \( (100 + 4) \). Then we use the identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = 100 \) and \( b = 4 \).
\( (100 + 4)³ = (100)³ + 3(100)²(4) + 3(100)(4)² + 4³ \)
\( = 1000000 + 3(10000)(4) + 300(16) + 64 \)
\( = 1000000 + 120000 + 4800 + 64 \)
\( = 1124864 \)
In simple words: We expanded these expressions by using the rule for cubing a sum of two terms. For numbers, we broke them into a sum of easier numbers first.

🎯 Exam Tip: When dealing with numerical cubes like \( (52)³ \), express them as a sum or difference of easily cubed numbers (e.g., \( (50+2)³ \) or \( (100-3)³ \)) to simplify calculations using algebraic identities.

Question 3. Expand
(i) \( (5 - x)³ \)
(ii) \( (2x - 4y)³ \)
(iii) \( (ab - c)³ \)
(iv) \( (48)³ \)
(v) \( (97xy)³ \)
Answer:
(i) To expand \( (5 - x)³ \), we use the identity \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Here, \( a = 5 \) and \( b = x \).
\( (5 - x)³ = 5³ - 3(5)²(x) + 3(5)(x²) - x³ \)
\( = 125 - 3(25)(x) + 15x² - x³ \)
\( = 125 - 75x + 15x² - x³ \)
(ii) To expand \( (2x - 4y)³ \), we use the identity \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Here, \( a = 2x \) and \( b = 4y \).
\( (2x - 4y)³ = (2x)³ - 3(2x)²(4y) + 3(2x)(4y)² - (4y)³ \)
\( = 2³x³ - 3(2²x²)(4y) + 3(2x)(4²y²) - (4³y³) \)
\( = 8x³ - 48x²y + 96xy² - 64y³ \)
(iii) To expand \( (ab - c)³ \), we use the identity \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Here, \( a = ab \) and \( b = c \).
\( (ab - c)³ = (ab)³ - 3(ab)²c + 3ab(c)² - c³ \)
\( = a³b³ - 3(a²b²)c + 3abc² - c³ \)
(iv) To find \( (48)³ \), we write \( 48 \) as \( (50 - 2) \). Then we use the identity \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Here, \( a = 50 \) and \( b = 2 \).
\( (50 - 2)³ = 50³ - 3(50)²(2) + 3(50)(2)² - 2³ \)
\( = 125000 - 15000 + 600 - 8 \)
\( = 110000 + 592 \)
\( = 110592 \)
(v) To find \( (97xy)³ \), we can first simplify it to \( (97)³ x³ y³ \). Now we calculate \( (97)³ \) by writing \( 97 \) as \( (100 - 3) \). Then we use the identity \( (a - b)³ = a³ - 3a²b + 3ab² - b³ \). Here, \( a = 100 \) and \( b = 3 \).
\( (100 - 3)³ = 100³ - 3(100)²(3) + 3(100)(3)² - 3³ \)
\( = 1000000 - 90000 + 2700 - 27 \)
\( = 910000 + 2673 \)
\( = 912673 \)
So, \( (97xy)³ = 912673x³y³ \)
In simple words: We used the rule for cubing a difference of two terms. For numbers like 48 and 97, we wrote them as a subtraction from a round number to make the calculation easier.

🎯 Exam Tip: Pay close attention to the signs when using the identity for \( (a-b)³ \), especially for the \( -3a²b \) and \( +3ab² \) terms. A common mistake is misplacing a negative sign.

Question 4. Simplify \( (p - 2)(p + 1)(p - 4) \)
Answer: To simplify \( (p - 2)(p + 1)(p - 4) \), we use the identity \( (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc \). Here, \( x = p \), \( a = -2 \), \( b = 1 \), and \( c = -4 \).
\( (p - 2)(p + 1)(p - 4) = (p + (-2))(p + 1)(p + (-4)) \)
\( = p³ + (-2 + 1 + (-4))p² + ((-2)(1) + (1)(-4) + (-4)(-2))p + (-2)(1)(-4) \)
\( = p³ + (-5)p² + (-2 - 4 + 8)p + 8 \)
\( = p³ - 5p² + 2p + 8 \)
In simple words: We multiplied three expressions together using a special algebraic rule. We added and multiplied the numbers correctly to get the final simplified answer.

🎯 Exam Tip: When simplifying products of three linear factors, remember the identity \( (x+a)(x+b)(x+c) \) and be careful with the signs when substituting values for a, b, and c.

Question 5. Find the volume of the cube whose side is \( (x + 1) \) cm
Answer: The side of the cube is given as \( (x + 1) \) cm. The volume of a cube is calculated by cubing its side.
Volume of the cube \( = (side)³ \)
\( = (x + 1)³ \) cm³
To expand \( (x + 1)³ \), we use the algebraic identity \( (a + b)³ = a³ + 3a²b + 3ab² + b³ \). Here, \( a = x \) and \( b = 1 \).
\( (x + 1)³ = x³ + 3x²(1) + 3x(1)² + 1³ \)
\( = x³ + 3x² + 3x + 1 \) cm³
In simple words: To find the space inside the cube, we take the length of one side and multiply it by itself three times. We used a math rule to expand the expression for the side.

🎯 Exam Tip: Clearly state the formula for the volume of a cube before applying the algebraic identity, and remember to include the correct units in the final answer.

Question 6. Find the volume of the cuboid whose dimensions are \( (x + 2),(x - 1) \) and \( (x - 3) \)
Answer: The dimensions of the cuboid are length \( (l) = (x + 2) \), breadth \( (b) = (x - 1) \), and height \( (h) = (x - 3) \). The volume of a cuboid is found by multiplying its length, breadth, and height.
Volume of the cuboid \( = l \times b \times h \)
\( = (x + 2)(x - 1)(x - 3) \) units³
To expand this product, we use the identity \( (x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc \). Here, \( a = 2 \), \( b = -1 \), and \( c = -3 \).
\( = x³ + (2 - 1 - 3)x² + (2(-1) + (-1)(-3) + (-3)(2))x + (2)(-1)(-3) \)
\( = x³ + (-2)x² + (-2 + 3 - 6)x + 6 \)
\( = x³ - 2x² - 5x + 6 \) units³
In simple words: We multiplied the three different side lengths of the cuboid to get its volume. We used a rule that helps us multiply three expressions together.

🎯 Exam Tip: Make sure to correctly identify 'a', 'b', and 'c' including their signs when using the \( (x+a)(x+b)(x+c) \) identity for cuboid volume calculations.

Objective Type Questions

Question 7. If \( x² - y² = 16 \) and \( (x + y) = 8 \) then \( (x - y) \) is
(a) 8
(b) 3
(c) 2
(d) 1
Answer: (c) 2
In simple words: We know that \( x² - y² \) can be written as \( (x + y)(x - y) \). Since we are given the values for \( x² - y² \) and \( x + y \), we can easily find \( x - y \) by dividing.

🎯 Exam Tip: Always remember the difference of squares identity, \( a² - b² = (a - b)(a + b) \), as it frequently helps simplify expressions and solve problems quickly.

Question 8. \( \frac{(a+b)(a^{3}-b^{3})}{(a^{2}-b^{2})} = \)
(a) \( a² - ab + b² \)
(b) \( a² + ab + b² \)
(c) \( a³ - b³ \)
(d) \( a² - 2ab + b² \)
Answer: (b) \( a² + ab + b² \)
In simple words: We used the rules for factoring (breaking down) \( a³ - b³ \) and \( a² - b² \). After writing them out, we could cancel the common terms to get the simpler answer.

🎯 Exam Tip: Master the factorization formulas for difference of cubes \( (a³-b³ = (a-b)(a²+ab+b²)) \) and difference of squares \( (a²-b² = (a-b)(a+b)) \) as they are essential for simplifying rational algebraic expressions.

Question 9. \( (p + q)(p² - pq + q²) \) is equal to
(a) \( p³ + q³ \)
(b) \( (p + q)³ \)
(c) \( p³ - q³ \)
(d) \( (p - q)³ \)
Answer: (a) \( p³ + q³ \)
In simple words: This is a direct match to a well-known algebraic identity. When you multiply these two expressions, the result is the sum of their cubes.

🎯 Exam Tip: Recognize the expansion of \( (a + b)(a² - ab + b²) \) as the formula for the sum of two cubes, \( a³ + b³ \), which is a fundamental algebraic identity.

Question 10. If \( (a - b) = 3 \) and \( ab = 5 \) then \( a³ - b³ = \)
(a) 15
(b) 18
(c) 62
(d) 72
Answer: (d) 72
In simple words: We used the formula for the difference of cubes, \( a³ - b³ = (a - b)³ + 3ab(a - b) \), and put in the given values for \( (a - b) \) and \( ab \) to find the answer.

🎯 Exam Tip: For problems involving \( a³ \pm b³ \) where \( (a \pm b) \) and \( ab \) are given, use the expanded form of \( (a \pm b)³ \) to derive the value without finding 'a' and 'b' separately.

Question 11. \( a³ + b³ = (a + b)³ \) minus
(a) \( 3a(a + b) \)
(b) \( 3ab(a - b) \)
(c) \( -3ab(a + b) \)
(d) \( 3ab(a + b) \)
Answer: (d) \( 3ab(a + b) \)
In simple words: The full identity for \( (a + b)³ \) is \( a³ + b³ + 3ab(a + b) \). If we want to find just \( a³ + b³ \), we move the \( 3ab(a + b) \) part to the other side by subtracting it.

🎯 Exam Tip: Understanding the relationship between \( (a+b)³ \) and \( a³+b³ \) allows you to quickly rearrange the identity \( (a+b)³ = a³+b³+3ab(a+b) \) to solve for either component.