Factorise $(x + 4)^{3} - 9 x - 36$ .

(x - 4) (x + 1) (x + 7)

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(x + 6) (x + 2) (x + 9)

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(x + 4) (x + 1) (x + 5)

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(x + 4) (x + 1) (x + 7)

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Solution

The correct option is C $(x + 4) (x + 1) (x + 7)$
We have,
$(x + 4)^{3} - 9 x - 36$
$= x^{3} + 64 + 12 x^{2} + 48 x - 9 x - 36 [∵ (a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}]$
$= x^{3} + 12 x^{2} + 39 x + 28$
By hit and trial $x + 1$ is one of the factor.
So, Divide $x^{3} + 12 x^{2} + 39 x + 28$ by $x + 1$ , we get
$x^{3} + 12 x^{2} + 39 x + 28 = (x + 1) (x^{2} + 11 x + 28)$
$\Rightarrow x^{3} + 12 x^{2} + 39 x + 28 = (x + 1) (x^{2} + 7 x + 4 x + 28)$
$\Rightarrow x^{3} + 12 x^{2} + 39 x + 28 = (x + 1) [x (x + 7) + 4 (x + 7)]$
$\Rightarrow x^{3} + 12 x^{2} + 39 x + 28 = (x + 1) (x + 4) (x + 7)$
So, $D$ is the correct option.

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