Using (a − b)³ = a³ − 3a²b + 3ab² − b³, expand (2x − 1)³.

8x³ − 12x² + 6x − 1. (2x)³ = 8x³; 3(2x)²(1) = 12x²; 3(2x)(1)² = 6x; 1³ = 1. For a difference the signs alternate: 8x³ − 12x² + 6x − 1. Note that (2x − 1)³ is not 8x³ − 1 — that would be the cube-difference identity.

Using $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ , expand $(2 x - 1)^{3}$ .

Class 9PolynomialsPolynomials

A $8 x^{3} + 12 x^{2} + 6 x + 1$
B $8 x^{3} - 12 x^{2} + 6 x - 1$
C $8 x^{3} - 1$
D $2 x^{3} - 12 x^{2} + 6 x - 1$

Step-by-Step Solution

(2 x)^{3} = 8 x^{3}

;

3 (2 x)^{2} (1) = 12 x^{2}

;

3 (2 x) (1)^{2} = 6 x

;

1^{3} = 1

. For a difference the signs alternate:

8 x^{3} - 12 x^{2} + 6 x - 1

. Note that

(2 x - 1)^{3}

is not

8 x^{3} - 1

— that would be the cube-difference identity.

Answer: B)

8 x^{3} - 12 x^{2} + 6 x - 1

Key idea: Factorising polynomials using the factor theorem

To factorise a cubic like

p (x) = x^{3} - 6 x^{2} + 11 x - 6

:

**Step 1**: Try small integer values:

p (1) = 1 - 6 + 11 - 6 = 0

. ✓ So

(x - 1)

is a factor.
**Step 2**: Divide

p (x)

(x - 1)

using long division. Quotient:

x^{2} - 5 x + 6

.
**Step 3**: Factor the quadratic:

x^{2} - 5 x + 6 = (x - 2) (x - 3)

.
**Step 4**: Combine:

p (x) = (x - 1) (x - 2) (x - 3)

.

The zeros are 1, 2, 3 — three real zeros, as expected for a cubic.

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Using (a−b)3=a3−3a2b+3ab2−b3, expand (2x−1)3.

Step-by-Step Solution

Key idea: Factorising polynomials using the factor theorem

Master this chapter in 30 minutes a day.

More Class 9 Polynomials practice

Using $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ , expand $(2 x - 1)^{3}$ .