# Thread: Work out the power

1. ## Work out the power

$64^{-\frac{2}{3}}$

No calculator

2. $64^{-\frac{2}{3}}$

$\frac{1}{64^{\frac{2}{3}}}$

$\frac{1}{\left(64^{\frac{1}{3}}\right)^2}$

$= \frac{1}{16}$

3. $64^{-\frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}}$

Now:

$64^{\frac{2}{3}} = \sqrt[3]{64}^2 = 4^2 = 16$

But we have the reciprocal so the answer is:

$\frac{1}{16}$

4. Originally Posted by rowe
$64^{-\frac{2}{3}} = \frac{1}{64^{\frac{2}{3}}}$

Now:

$64^{\frac{2}{3}} = \sqrt[3]{64}^2 = 4^2 = 16$

But we have the reciprocal so the answer is:

$\frac{1}{16}$
Originally Posted by bigwave
$64^{-\frac{2}{3}}$

$\frac{1}{64^{\frac{2}{3}}}$

$\frac{1}{\left(64^{\frac{1}{3}}\right)^2}$

$= \frac{1}{16}$
Thank you but I prefer bigwave's method. Thanks bigwave!

5. Actually. Both are hard without a calculator :/

6. Originally Posted by Mukilab
Actually. Both are hard without a calculator :/
The key to this question is knowing/working out that $64 = 4^3$

If you prefer you can use prime factors

$64 = 2 \times 32$
$32 = 2 \times 16$
$16 = 2 \times 8$
$8 = 2 \times 4$
$4 = 2 \times 2$

Therefore $64 = 2^6 = 2^{3+3} = 2^3 \cdot 2^3$

$(\sqrt[3]{64})^2 = [\sqrt[3]{2^3} \times \sqrt[3]{2^3}]^2$

Cubes and cube roots cancel to leave $(2 \times 2)^2 = 16
$

7. Oh ok. Simplified a complicated looking question ^^