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

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Answer 1 · 2018-04-09T22:22:05

#((sqrt(6)+sqrt(2))/4)^6+((sqrt(6)-sqrt(2))/4)^6=13/16#

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)#

#=2sqrt(2+sqrt(3))#

Likewise, for the other numerator, we can write

#sqrt(6)-sqrt(2)=2sqrt(2-sqrt(3))#

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

#((2+sqrt(3))^(1/2)/2)^6+((2-sqrt(3))^(1/2)/2)^6#

#(2+sqrt(3))^3/2^6+(2-sqrt(3))^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!)

#(8+3*4sqrt(3)+3*2*3+3^(3/2))/64+(8-3*4sqrt(3)+3*2*3-3^(3/2))/64#

Which simplifies to

#(16+36)/64=13/16#