# How do you solve 63= - 1+ k ^ { \frac { 3} { 2} }?

Feb 21, 2017

Change ${k}^{\frac{3}{2}}$ to $k$

$k = 16$

#### Explanation:

${k}^{\frac{3}{2}}$ can be written as ${\sqrt[2]{k}}^{3}$

$63 = - 1 + {k}^{\frac{3}{2}}$

$63 + 1 = {k}^{\frac{3}{2}}$

$64 = {k}^{\frac{3}{2}}$

To remove the cube take the cube root of 64, which is 4.

root3 64 = root3 (k^(3/2)

$4 = {k}^{\frac{1}{2}}$

And then to remove the power of $\frac{1}{2}$ from $k$ you take

${4}^{2}$ and ${\left({k}^{\frac{1}{2}}\right)}^{2}$

$16 = k$