Ab-surder!

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 $\frac{\sqrt[3]{4}}{\bigl(\sqrt[3]{2}\bigr)^3}=\frac{\sqrt[3]{4}}{2}.$

Polynomials & Rational Functions

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