How do you factor $(x^{3}) (x^{6}) + (8 z^{3})$ $(x^3)(x^6)+(8z^3)$ ?

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Answer 1 · 2015-05-06T23:32:21

Note the general form for the factoring of a sum of cubes:
$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3+b^3 = (a+b)(a^2-ab+b^2)$

$(x^{3}) (x^{6}) + (8 z^{3})$ $(x^3)(x^6) + (8z^3)$

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

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

How do you factor (x3)(x6)+(8z3)(x^3)(x^6)+(8z^3)?