value of $x^3 - 6x^2 + 6x$ when $x$ is an expression involving radicals

Question

If $x = 2 + 2^{\frac13} + 2^{\frac23}$, then what is the value of $x^3 - 6x^2 + 6x$?

How would I solve it? Surely plugging in is one way, but there's got to be some other way.

Why? Plugging in is a simple exercise. Of course, you can factor out x — M.B., Apr 04 '14 at 05:23
@M.B.The source of this questions had a lot of difficult and tricky questions and this question is the last one...so there has to be some clever way of doing it... — Shaurya Gupta, Apr 04 '14 at 05:25
@M.B. I have tried factoring out x and then plugging in...the answer comes 2, but it's quite boring — Shaurya Gupta, Apr 04 '14 at 05:33
Or.. Did not help that much, but it simplifies somewhat to write the sum of 2s as the sum of a geometric series. — M.B., Apr 04 '14 at 05:47
Another way is to use field theory. See https://math.stackexchange.com/a/4811203/589 — lhf, Nov 21 '23 at 14:50

score 3 · Accepted Answer · answered Apr 04 '14 at 06:01

3

$x=2+2^{\frac13}+2^{\frac23}\implies (x-2)^3=(2^{\frac13}+2^{\frac23})^3$

$\implies x^3-8-6x^2+12x=2+4+6(x-2)=6x-6$

$\implies x^3-6x^2+6x=2$

answered Apr 04 '14 at 06:01

Aang

14,672

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And that's why I hate plugging in... – Shaurya Gupta Apr 04 '14 at 06:04

score 0 · Answer 2 · edited Apr 04 '14 at 06:05

0

You can factor it as $2(x-1)^3 - x^3 + 2$. Since $(x-1)\left(2^{1/3}-1\right) = 1$ and $x\left(1-2^{-1/3}\right) = 1$ we get $$\left(\frac{2^{1/3}}{2^{1/3}-1}\right)^3 - \left(\frac{1}{1-2^{-1/3}}\right)^3 + 2.$$ So the left two terms cancel out.

edited Apr 04 '14 at 06:05

MJD

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answered Apr 04 '14 at 05:52

arando

109

value of $x^3 - 6x^2 + 6x$ when $x$ is an expression involving radicals

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