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How can I rationalize the following equation:

$\frac{22}{4\sqrt[3]{9}+2\sqrt[3]{6}+\sqrt[3]{4}}$

1 Answers 1

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Let $a=2\sqrt[3]{3}, b=\sqrt[3]{2}$. Then this is equal to $\dfrac{22}{a^2+ab+b^2} = \dfrac{22(a-b)}{a^3-b^3}$.

Since $a^3-b^3 = 8(3)-2 =22$, we have that our desired answer is $\dfrac{22(a-b)}{a^3-b^3} = a-b = 2\sqrt[3]{3}-\sqrt[3]{2}$.

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    This one really requires a bit more attention than most. Nicely done.2012-12-16