How do you factor completely a^3-10?

Jul 13, 2017

${a}^{3} - 10 = \left(a - \sqrt[3]{10}\right) \left({a}^{2} + a \sqrt[3]{10} + {\sqrt[3]{10}}^{2}\right)$
$10$ is not a perfect cube, but we can show its cube root as $\sqrt[3]{10}$ which is an irrational number.
${x}^{3} - {y}^{3} = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$
${a}^{3} - 10 = \left(a - \sqrt[3]{10}\right) \left({a}^{2} + a \sqrt[3]{10} + {\sqrt[3]{10}}^{2}\right)$