How to factor $16 x^{9}$ $16x^9$ + $54 y^{3}$ $54y^3$ using difference of cubes?

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How to factor $16 x^{9}$ $16x^9$ + $54 y^{3}$ $54y^3$ using difference of cubes?

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Answer 1 · 2017-10-10T06:13:42

$16 x^{9} + 54 y^{3} = 2 (2 x^{3} + 3 y) (4 x^{4} - 6 x^{3} y + 9 y^{2})$ $16x^9+54y^3=2(2x^3+3y)(4x^4-6x^3y+9y^2)$

Explanation:

Here we have sum of cubes and not difference of cubes. Hence, we have to use $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3=(a+b)(a^2-ab+b^2)$ . As such

$16 x^{9} + 54 y^{3}$ $16x^9+54y^3$

= $2 (8 x^{9} + 27 y^{3})$ $2(8x^9+27y^3)$

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

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

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

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