Using a suitable identity, find the value of (−12)³ + 7³ + 5³ given that (−12) + 7 + 5 = 0.

-1260. Since (−12) + 7 + 5 = 0, the identity gives a³ + b³ + c³ = 3abc = 3 × (−12) × 7 × 5 = 3 × (−420) = −1260.

Using a suitable identity, find the value of $(- 12)^{3} + 7^{3} + 5^{3}$ given that $(- 12) + 7 + 5 = 0$ .

Class 9PolynomialsPolynomials

Since

(- 12) + 7 + 5 = 0

, the identity gives

a^{3} + b^{3} + c^{3} = 3

ab c = 3 \times (- 12) \times 7 \times 5 = 3 \times (- 420) = - 1260

Answer: C) -1260

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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