# How do you simplify ( \frac { \sqrt { 6} + \sqrt { 2} } { 4} ) ^ { 6} + ( \frac { \sqrt { 6} - \sqrt { 2} } { 4} ) ^ { 6}?

Apr 9, 2018

${\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)}^{6} + {\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)}^{6} = \frac{13}{16}$

#### Explanation:

I'm going to do this in a tricky way, but I hope it will save us some work.

First, let's see if we can reduce the power of the first numerator by rewriting it as a NESTED radical.

sqrt(6)+sqrt(2)=sqrt((sqrt(6)+sqrt(2))^2)=sqrt(6+2+2sqrt(12)

$= 2 \sqrt{2 + \sqrt{3}}$

Likewise, for the other numerator, we can write

$\sqrt{6} - \sqrt{2} = 2 \sqrt{2 - \sqrt{3}}$

Now we can substitute these expressions into the numerators and focus on simplifying

${\left({\left(2 + \sqrt{3}\right)}^{\frac{1}{2}} / 2\right)}^{6} + {\left({\left(2 - \sqrt{3}\right)}^{\frac{1}{2}} / 2\right)}^{6}$

${\left(2 + \sqrt{3}\right)}^{3} / {2}^{6} + {\left(2 - \sqrt{3}\right)}^{3} / {2}^{6}$

Now we can write out the binomial expansion of a cubed expression (much easier than writing out the expansion of something taken to the 6th power!)

$\frac{8 + 3 \cdot 4 \sqrt{3} + 3 \cdot 2 \cdot 3 + {3}^{\frac{3}{2}}}{64} + \frac{8 - 3 \cdot 4 \sqrt{3} + 3 \cdot 2 \cdot 3 - {3}^{\frac{3}{2}}}{64}$

Which simplifies to

$\frac{16 + 36}{64} = \frac{13}{16}$