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Cube Root Using Prime Factorisation

Factorize:8x3...

Question

Factorize:

$8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$

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Solution

Given: $8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$

$∴ 8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$

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

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

$[∵ a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} = (a + b)^{3}]$

$= (2 x + y) (2 x + y) (2 x + y)$

Hence, $8 x^{3} + y^{3} + 12 x^{2} y + 6 x y^{2}$

$= (2 x + y) (2 x + y) (2 x + y) .$

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Cube Root Using Prime Factorisation

Standard VIII Mathematics

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