Polynomials & Rational Functions
Go and think about it...

Ab-surder!
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Warm-up

In Ab-surd!, we explored rationalising the denominators of fractions with square roots.

What would happen if instead of square roots, we had cube roots?

Can you rationalise the denominator of \(\dfrac{1}{\sqrt[3]{2}}\)?

If we multiply both the numerator and denominator by \(\bigl(\sqrt[3]{2}\bigr)^2=\sqrt[3]{4}\), we get \[\frac{\sqrt[3]{4}}{\bigl(\sqrt[3]{2}\bigr)^3}=\frac{\sqrt[3]{4}}{2}.\]

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Last updated 01-Mar-17
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