5

Calculate $$S=3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}-\sqrt[3]{2}-\sqrt[3]{20}+\sqrt[3]{25}$$ $\color{red}{\text{without using calculator}.}$

Please help me, I can't find any solution to sovle it.

NoChance
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mja
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  • Are you looking for an exact (algebraic) solution? Why do you think that there should be one? Is an approximate (numeric) solution acceptable? For the last two terms $\sqrt[3]{25}-\sqrt[3]{20}=\sqrt[3]5(\sqrt[3]5-\sqrt[3]4)$, which matches the first radical. – Ross Millikan Aug 13 '15 at 14:25
  • @RossMillikan I'm looking for an exact solution. Let $a=\sqrt[3]5, b=\sqrt[3]2$ I have $S=3\sqrt{a-b^2}+a(a-b^2)-b$. But I don't know how to continue. – mja Aug 13 '15 at 14:33
  • I don't think you will get a solution without using $5$ and $4$. Alpha says it is $0$, so we can multiply by conjugates as we wish. I think we need to find something clever to do with the $\sqrt[3]2$ – Ross Millikan Aug 13 '15 at 14:45
  • I had good fund with your question! :-) – Math-fun Aug 13 '15 at 18:43

2 Answers2

10

Let

$$s=\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}$$

If you carefully square $s$, you find

$$s^2=9(\sqrt[3]5-\sqrt[3]4)$$

Note that $\sqrt[3]2\gt1$, $\sqrt[3]{20}\gt2$, and $\sqrt[3]{25}\lt3$, so $s\gt0$. Thus $s=3\sqrt{\sqrt[3]5-\sqrt[3]4}$, hence $S=0$.

Barry Cipra
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  • Thanks, but how we know to let $s=\sqrt[3]2+\sqrt[3]{20}-\sqrt[3]{25}$ – mja Aug 13 '15 at 15:25
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    That's a definition. The Idea here is to look at this term more closely, and to help doing that, you just give it a name – in this case, $s$ was chosen. – Lukas Juhrich Aug 13 '15 at 15:31
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    @DDK, I'd like to call it inspiration, but it's really just experience. The key idea was to get rid of the square root, so it made sense to move those little cube roots over to the left hand side. I didn't feel like writing them all down, so I called them $s$. That way I knew the left hand side, when squared, would be $S^2-2sS+s^2$. – Barry Cipra Aug 13 '15 at 15:31
  • Thanks everyone for sharing your experience. :) – mja Aug 13 '15 at 15:35
1

\begin{align} (9+4\sqrt[3]{10})(5-\sqrt[3]{10^2})&=5+20\sqrt[3]{10}-9\sqrt[3]{10^2}\\ &=25-20+20\sqrt[3]{10}-9\sqrt[3]{10^2}\\ &=5^2-2\sqrt[3]{10}\sqrt[3]{10^2}+20\sqrt[3]{10}-9\sqrt[3]{10^2}\\ &=5^2-2\sqrt[3]{10}\sqrt[3]{10^2}+20\sqrt[3]{10}-2\times5\sqrt[3]{10^2}+\sqrt[3]{10^2}\\ &=5^2-2\sqrt[3]{10}\sqrt[3]{10^2}+\sqrt[3]{10^4}+2\times5\sqrt[3]{10}-2\times5\sqrt[3]{10^2}+\sqrt[3]{10^2}\\ &=(5-\sqrt[3]{10^2}+\sqrt[3]{10})^2 \end{align} Hence $$(9+4\sqrt[3]{10})(20-4\sqrt[3]{10^2})=2^2(5-\sqrt[3]{10^2}+\sqrt[3]{10})^2$$ therefore \begin{align} \Big(\sqrt{9+4\sqrt[3]{10}}-\sqrt{20-4\sqrt[3]{10^2}}\Big)^2&=(9+4\sqrt[3]{10})+(20-4\sqrt[3]{10^2})-2\sqrt{(9+4\sqrt[3]{10})(20-4\sqrt[3]{10^2})}\\ &=9 \end{align} or $$-3+\sqrt{9+4\sqrt[3]{10}}=\sqrt{20-4\sqrt[3]{10^2}}$$ or more simply with $\sqrt{20-4\sqrt[3]{10^2}}=\sqrt{\Big(2\sqrt[3]{5}\Big)^2\Big(\sqrt[3]{5}-\sqrt[3]{4}\Big)}$ we have $$\frac{-3+\sqrt{9+4\sqrt[3]{10}}}{2\sqrt[3]{5}}=\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$$

Finally let $u=\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}$ and define $$f(u)=3u-\sqrt[3]{2}+\sqrt[3]{5}u^2$$with $\displaystyle u_1=\frac{-3+\sqrt{9+4\sqrt[3]{2}\sqrt[3]{5}}}{2\sqrt[3]{5}}$ as a root.

Math-fun
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