How do you factor 2x^3-162?

Jul 1, 2015

$2 {x}^{3} - 162 = 2 \left({x}^{3} - {3}^{4}\right) = 2 \left({x}^{3} - {\left({3}^{\frac{4}{3}}\right)}^{3}\right)$

$= 2 \left(x - {3}^{\frac{4}{3}}\right) \left({x}^{2} + {3}^{\frac{4}{3}} x + {3}^{\frac{8}{3}}\right)$

$= 2 \left(x - 3 \sqrt[3]{3}\right) \left({x}^{2} + 3 \sqrt[3]{3} x + 9 \sqrt[3]{9}\right)$

Explanation:

Using the difference of cubes identity:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = x$ and $b = {3}^{\frac{4}{3}} = 3 \sqrt[3]{3}$

I suspect a typo in the problem as stated, perhaps it should have been:

$2 {x}^{4} - 162$

which would factor much more nicely.