How do you factor $y= 3x^3 + 81$ ?

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Answer 1 · 2018-03-06T13:36:53

$y=3(x+3)(x^2+9-3x)$

$y=3x^3+81$

$=>y=3(x^3+27)$

$=>y=3(x^3+3^3)$

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

So , $y=3(x^3+3^3)$

$=> y=3(x+3)(x^2+3^2-x xx 3)$

$=> y=3(x+3)(x^2+9-3x)$

How do you factor y= 3x^3 + 81 y= 3x^3 + 81 ?