How do you factor $2 x^{3} - 162$ $2x^3-162$ ?

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Answer 1 · 2015-07-01T11:09:57

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

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

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

Using the difference of cubes identity:

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

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

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

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

which would factor much more nicely.

How do you factor 2x^3-1622x3−1622x^3-162?