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Mean

Factorize: 27...

Question

Factorize: $27 p^{3} + 8 q^{3}$

A

(3 p + 2 q) [9 p^{2} - 6 p q + 4 q^{2}]

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B

(2 p + 3 q) [9 p^{2} - 6 p q + 4 q^{2}]

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C

(3 p + 2 q) [8 p^{2} - 4 p q + 4 q^{2}]

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D

(3 p + 2 q) [6 p^{2} - 6 p q + 2 q^{2}]

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Solution

The correct option is A $(3 p + 2 q) [9 p^{2} - 6 p q + 4 q^{2}]$
$27 p^{3} + 8 q^{3}$

$= 3^{3} p^{3} + 2^{3} q^{3}$

$= (3 p)^{3} + (2 q)^{3}$

Comparing this with $a^{3} + b^{3}$ ,
we get a = 3p, b = 2q

Using the identity, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$(3 p)^{3} + (2 q)^{3} = (3 p + 2 q) [(3 p)^{2} - (3 p) (2 q) + (2 q)^{2}]$
$27 p^{3} + 8 q^{3} = (3 p + 2 q) [9 p^{2} - 6 p q + 4 q^{2}]$

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