# Thread: How are these 'the same'?

1. ## How are these 'the same'?

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

Stuck again on this one.
Any help would be appreciated.

2. Yuck. I started with the left-hand side.

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

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

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

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

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

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

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

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.

3. Thank you!