# How do you simplify 6root4 9 +2 root3 9 +2 root4 9 - 3 root3 9?

Mar 29, 2016

$\sqrt{3} \left(\pm 6 \pm 2\right) - {9}^{\frac{1}{3}}$ = sqrt3(+-6+-2)-2.08), nearly.
The four values are $11.78 , 4.85 , - 9.01 \mathmr{and} - 15.94$, nearly.

#### Explanation:

${9}^{\frac{1}{4}} = {\left({9}^{\frac{1}{2}}\right)}^{\frac{1}{2}} = {\left(+ 3\right)}^{\frac{1}{2}} = \pm \sqrt{3}$.

The first and third tems add up to $\sqrt{3} \left(\pm 6 \pm 2\right)$.

The second and fourth add up to -9^(1/3..

So, the expression simplifies to $\sqrt{3} \left(\pm 6 \pm 2\right) - {9}^{\frac{1}{3}}$.
There is only one real value 2.08, nearly, for #9^(1/3).

In general, ${a}^{\frac{1}{n}}$ has n values. The complex values occur in conjugate pairs. For example, the values ${1}^{\frac{1}{3}}$are $1 , \pm i$.,