Class 8 Maths Chapter 8 Question Answers - Algebraic Expressions and Identities

Q1: Simplify the following:
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
Ans: 
(i) a² (b² – c²) + b² (c² – a²) + c² (a² – b²)
= a²b² – a²c²) + b²c² – b²a²) + c²a² – c²b²)
= 0
(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)
= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³
= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³
= 6x³ – x²y² – 2xy³

Q2: Multiply (6x² – 5x + 3) by (3x² + 7x – 3)
Ans: (6x² – 5x + 3) × (3x² + 7x – 3)
= 6x²(3x² + 7x – 3) – 5x(3x² + 7x – 3) + 3(3x² + 7x – 3)
= 18x⁴ + 42x³ – 18x² – 15x³ – 35x² + 15x + 9x² + 21x – 9
= 18x⁴ + 42x³ – 15x³ – 18x² – 35x² + 9x² + 15x + 21x – 9
= 18x⁴ + 27x³ – 44x² + 36x – 9

Q3:  Multiply (3x² + 5y²) by (5x² – 3y²)
Ans: (3x² + 5y²) × (5x² – 3y²)
= 3x²(5x² – 3y²) + 5y²(5x² – 3y²)
= 15x⁴ – 9x²y² + 25x²y² – 15y⁴
= 15x⁴ + 16x²y² – 15y⁴

Q4: Multiply x² + 2y by x³ – 2xy + y³ and find the value of the product for x = 1 and y = -1.

Ans: (x² + 2y) × (x³ – 2xy + y³)
= x²(x³ – 2xy + y³) + 2y(x³ – 2xy + y³)
= x⁵ – 2x³y + x²y³ + 2x³y – 4xy² + 2y⁴
= x⁵ + x²y³ – 4xy² + 2y⁴
Put x = 1 and y = -1
= (1)⁵ + (1)² (-1)³ – 4(1)(-1)² + 2(-1)⁴
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

Q5: Find the value of x, if 10000x = (9982)² – (18)² 
Ans: RHS = (9982)² – (18)² = (9982 + 18)(9982 – 18)
[Since a² -b² = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964

Q6: Simplify: 2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)
Ans: 2x²(x + 2) – 3x(x² – 3) – 5x(x + 5)
= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x
= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x
= -x³ – x² – 16x

Q7: Verify that (11pq + 4q)² – (11pq – 4q)² = 176pq² 
Ans:  LHS = (11pq + 4q)² – (11pq – 4q)² = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a² -b² = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq²
= RHS.
Hence Verified.