How do you factor completely $a^{3} - 10$ $a^3-10$ ?

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Answer 1 · 2017-07-13T07:09:35

$a^{3} - 10 = (a - \sqrt[3]{10}) (a^{2} + a \sqrt[3]{10} + {\sqrt[3]{10}}^{2})$ $a^3 -10 = (a-root3 10)(a^2+aroot3 10 +root3 10^2)$

$10$ $10$ is not a perfect cube, but we can show its cube root as $\sqrt[3]{10}$ $root3 10$ which is an irrational number.

The difference of cubes factorises as:
$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ $x^3 -y^3 = (x-y)(x^2 +xy+y^2)$

$a^{3} - 10 = (a - \sqrt[3]{10}) (a^{2} + a \sqrt[3]{10} + {\sqrt[3]{10}}^{2})$ $a^3 -10 = (a-root3 10)(a^2+aroot3 10 +root3 10^2)$

How do you factor completely a3−10a^3-10?