Show that $\displaystyle (\sqrt[3]{\frac{1}{16}})^{-1} + 8(\sqrt[3]{\frac{1}{16}})^2 = 3*2^\frac{1}{3}.$

Stuck again on this one.

Any help would be appreciated.

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- Nov 24th 2010, 01:16 PM #1

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- Nov 24th 2010, 01:42 PM #2
Yuck. I started with the left-hand side.

$\displaystyle (\sqrt[3]{\frac{1}{16}})^{-1} + 8(\sqrt[3]{\frac{1}{16}})^2 = 3*2^\frac{1}{3}$

The left hand side$\displaystyle =((2^{-4})^{\frac{1}{3}})^{-1} + 2^3((2^{-4})^{\frac{1}{3}})^2$

Apply the $\displaystyle (x^a)^b =x^{a \times b}$ and $\displaystyle (x^a)(x^b)=x^{a+b} $rules.

$\displaystyle =2^{(\frac{4}{3})}+2^{\frac{-8}{3}+3}$

$\displaystyle =2^{\frac{4}{3}}+2^\frac{1}{3}}$

$\displaystyle =2^{\frac{1}{3}}(2^1+1)$

$\displaystyle =3(2^{\frac{1}{3}})$

I'm sure after my hours of work, someone will have already provided a much neater solution, but this is how I would approach the question. There is probably a much easier method, but this works nicely for me.

- Nov 24th 2010, 01:49 PM #3