2006 Cyprus MO/Lyceum/Problem 9

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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.

See also

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