2006 Cyprus MO/Lyceum/Problem 9

Problem

If $x=\sqrt[3]{4}$ and $y=\sqrt[3]{6}-\sqrt[3]{3}$ , then which of the following is correct?

$\mathrm{(A)}\ x=y\qquad\mathrm{(B)}\ x<y\qquad\mathrm{(C)}\ x=2y\qquad\mathrm{(D)}\ x>2y\qquad\mathrm{(E)}\ \text{None of these}$

Solution

The question is asking us for an approximation of the ratio between $x : y$ . Thus, we are allowed to multiply both sides by a constant.

By difference of cubes, $\begin{align*}\sqrt[3]{4}(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})&:(\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\ 2\sqrt[3]{18}+2\sqrt[3]{9}+\sqrt[3]{36}&:3\end{align*}$ We can approximate the terms on the LHS; $2\sqrt[3]{18} > 4$ , $2\sqrt[3]{9} > 4$ , $\sqrt[3]{36} > 3$ , so the sum on the left side $> 11$ . Hence $x > 2y$ , and the answer is $\mathrm{(D)}$ .

Remark: There doesn't seem to be any direct way to calculate a simple ratio between the two terms, but various variations can involve approximating terms by multiplying by certain quantities.

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 8	Followed by Problem 10
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2006 Cyprus MO/Lyceum/Problem 9

Problem

Solution

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