Proving an expression is $4$

Question

Is $\displaystyle\sqrt[\huge3]{\frac{1}{2} \left(56-\sqrt{\frac{84640}{27}}\right)}+\sqrt[\huge 3]{\frac{1}{2} \left(\sqrt{\frac{84640}{27}}+56\right)}=4$ true ?

This was asked during an oral examination where calc and CAS are forbidden.

Mathematica seems to say it's true.

Can someone find a nice proof ?

cube both sides and you will get the result. – Satvik Mashkaria Jun 08 '14 at 11:04 — Satvik Mashkaria, Jun 08 '14 at 11:04

Git Gud · Accepted Answer · 2014-06-08T15:50:59.263

3

Let $\alpha =\dfrac {56}2, \beta=\dfrac 1 2\sqrt{\dfrac{84640}{27}}$ and $\gamma =\root 3\of {\alpha -\beta}+\root 3\of {\alpha +\beta}$.

All the numbers taken are supposed to be real.

Now note that $$\begin{align} \gamma ^3&=\left(\root 3\of {\alpha -\beta}\right)^3+3\root 3\of {\alpha -\beta}^2\root 3\of {\alpha +\beta}+3\root 3\of {\alpha -\beta}\root 3\of {\alpha +\beta}^2+\left(\root 3\of {\alpha +\beta}\right)^3\\ &=\alpha -\beta +3\root 3\of {\alpha -\beta}\root 3\of {\alpha +\beta}\Bigl(\underbrace{\root 3\of {\alpha -\beta}+\root 3\of {\alpha +\beta}}_{\huge =\gamma}\Bigr)+\alpha +\beta\\ &=2\alpha+3\root 3\of {\alpha ^2-\beta ^2}\gamma. \end{align}$$

Since $\alpha ^2=784$ and $\beta ^2=\dfrac{21160}{27}$ it follows that $\alpha ^2-\beta ^2=\dfrac{21168}{27}-\dfrac{21160}{27}=\dfrac 8{27}$.

Thus $\root 3\of {\alpha ^2+\beta ^2}=\dfrac 2 3$.

One concludes that $\gamma$ is such that $\gamma^3=56+2\gamma$, hence $\gamma$ is a (the only) real root of $x^3-2x-56$.

edited Jun 08 '14 at 15:50

answered Jun 08 '14 at 11:27

Git Gud

31,356

There's a slight mistake in the first and second that follow $\gamma^3$ – Gabriel Romon Jun 08 '14 at 11:29
@G.T.R I can't understand your comment. Can you tell me in which line the mistake is? – Git Gud Jun 08 '14 at 11:30
I edited it since it was really minor. – Gabriel Romon Jun 08 '14 at 11:32
@G.T.R Yes, yes. You're right. Too much copy and paste. Thank you. – Git Gud Jun 08 '14 at 11:33
Nice and clear solution, thank you! – Gabriel Romon Jun 08 '14 at 11:35
@G.T.R You're welcome. I tend to go the extra mile for regular users. – Git Gud Jun 08 '14 at 11:36

score 3 · Answer 2 · answered Jun 08 '14 at 12:58

What are these guys sadists ? How is one supposed to answer this in an oral exam ? Anyway here is another solution, we recognise the expression as the form of Cardano's solution to the cubic $x^3+px+q=0$ $$\frac{1}{3}\left( \sqrt[3]{-\frac{27}{2}q + \frac{3\sqrt{-3D}}{2}}+ \sqrt[3]{-\frac{27}{2}q - \frac{3\sqrt{-3D}}{2}} \right)$$ where $D=-27q^2-4p^3$ if we multiply the $\frac{1}{3}$ through we have $$ \sqrt[3]{\frac{1}{2}(-q + \sqrt{ \frac{-3D}{27} })}+ \sqrt[3]{\frac{1}{2}(q - \sqrt{\frac{-D}{27}}} )$$ So we recognise immediately that $q=-56$ and $-D=84640$ so $$27q^2+4p^3=84640$$ and this gives $p=-2$ So the cubic in question is $x^3-2x-56$ which has the root $4$.

score 0 · Answer 3 · answered Jun 08 '14 at 11:12

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cubing both sides, $56+(\frac{1}{4}(56^2-\frac{84640}{27}) )^{1/3}(4)=64$ simplify and you will get the result.

answered Jun 08 '14 at 11:12

Satvik Mashkaria

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Proving an expression is $4$

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