How do you factor the expression $16 x^{3} + 54 y^{6}$ $16x^3 + 54y^6$ ?

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Answer 1 · 2015-12-20T18:27:52

Factor out a $2$ $2$ and apply the sum of cubes formula to find that

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

The sum of cubes formula states that

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

(verify this by expanding the right hand side)

With this available, we have

$16 x^{3} + 54 y^{6} = 2 (8 x^{3} + 27 y^{6})$ $16x^3 + 54y^6 = 2(8x^3 + 27y^6)$

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

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

(applying the sum of cubes formula)

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

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

How do you factor the expression 16x3+54y616x^3 + 54y^6?