Factorise the following: $y^{6} + 32 y^{3} - 64$

(y + 2) (y^{4} - 2 y^{3} + 8 y^{3} + 8 y + 16)

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(y^{2} + 2 y - 4) (y^{4} - 2 y^{3} + 8 y^{2} + 8 y + 16)

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(y^{2} + 2 y - 4) (y^{4} - 2 y^{3} + 8 y^{2})

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(y + 2) (y^{4} - 2 y^{3} + 8 y^{2} + 8 y)

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Solution

The correct option is C $(y^{2} + 2 y - 4) (y^{4} - 2 y^{3} + 8 y^{2} + 8 y + 16)$
$y^{6} + 32 y^{3} - 64$
$= y^{6} + 8 y^{3} - 64 + 24 y^{3}$
$= (y^{2})^{3} + (2 y)^{3} - 4^{3} - 3 (y^{2}) (2 y) (- 4)$
Using, $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - a c - b c)$
$= (y^{2} + 2 y - 4) ((y^{2})^{2} + (2 y)^{2} + (- 4)^{2} - (y^{2}) (2 y) - (y^{2}) (- 4) - (2 y) (- 4))$
$= (y^{2} + 2 y - 4) (y^{4} + 4 y^{2} + 16 - 2 y^{3} + 4 y^{2} + 8 y)$
$= (y^{2} + 2 y - 4) (y^{4} + 8 y^{2} + 16 - 2 y^{3} + 8 y)$

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