Given: f(x)=27x3+8 =3x3+23 [27x3=33x3=3x3] =3x+23x2-3x×2+22 [a3+b3=a+ba2-ab+b2] =(3x+2)9x2-6x+4

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Factorise: $27 x^{3} + 8$ ?

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(3 x + 2) (9 x^{2} - 6 x + 4)

(3 x - 2) (9 x^{2} + 6 x - 4)

(3 x + 2) (9 x^{2} + 6 x + 4)

(3 x + 2) (9 x^{2} - 6 x - 4)

answer is A.

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Detailed Solution

Given:

f (x) = 27 x^{3} + 8

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

[

{27 x}^{3} = (3^{3} x^{3}) = {(3 x)}^{3}

]

= (3 x + 2) \{{(3 x)}^{2} - 3 x \times 2 + 2^{2}\}

[

{a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})}^{}

]

= (3 x + 2) (9 x^{2} - 6 x + 4)

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