If $x = 2 + 2^{2/3} + 2^{1/3}$, then find the value of $f(x)=x^3 - 6x^2 + 6x$.
I am unable to get to the answer - end up with more than one term. Please help me solve this!
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3May be answer contain more than one term. – Bumblebee Sep 16 '14 at 09:04
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@Nilan - actually buddy i knew the answer was 2 :) – Mohit Singh Sep 16 '14 at 09:32
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Already http://math.stackexchange.com/questions/739174/value-of-x3-6x2-6x-when-x-is-an-expression-involving-radicals – Macavity Sep 16 '14 at 18:54
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@Macavity Oops! I should have performed a thorough search. Thanks for pointing it out friend. – Mohit Singh Sep 16 '14 at 19:49
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Letting $t=2^{1/3}\Rightarrow t^3=2$, we have $$(x-2)^3=(t^2+t)^3$$$$\Rightarrow x^3-6x^2+12x-8=t^6+3t^5+3t^4+t^3=4+6t^2+6t+2.$$ Then, we have $$\begin{align}x^3-6x^2+6x&=(6t^2+6t+6)+8-6x\\&=6t^2+6t+14-6(t^2+t+2)\\&=14-12\\&=2.\end{align}$$
mathlove
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We have $$x-2=2^{\frac23}+2^{\frac13}$$
Cubing we get, $$x^3-3x^2(2)+3x(2^2)+2^3=(2^{\frac23})^3+(2^{\frac13})^3+3\cdot2^{\frac23}\cdot2^{\frac13}(2^{\frac23}+2^{\frac13})$$
$$\iff x^3-6x^2+12x+8=2+2^2+6(x-2)$$
Can you take it home from here?
lab bhattacharjee
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Hey buddy I guess you mistakenly copied the question wrong - it is x-2 = 2^2/3 + 2^1/3 and not "2^2/3 + 2^2/3". Thanks for taking interest though! :) – Mohit Singh Sep 16 '14 at 11:36
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@MohitSingh, Hope you understood the typo which has been rectified now – lab bhattacharjee Sep 16 '14 at 11:38