```Question 174247
Is the expression {{{root(6,sqrt(2))-root(12,2^13)}}} ???

{{{root(6,sqrt(2))-root(12,2^(12+1))}}} Break up {{{2^(13)}}} to {{{2^(12+1)}}}

{{{root(6,sqrt(2))-root(12,2^(12)*2^(1))}}} Break up the exponent (see note below)

{{{root(6,sqrt(2))-root(12,2^(12))*root(12,2^(1)))}}} Break up the square root (see note below).

{{{root(6,sqrt(2))-2*root(12,2)}}} Take the 12th root of {{{2^12}}} to get 2 (see note below).

{{{root(6,2^(1/2))-2*root(12,2)}}} Convert from radical notation to exponential notation. So {{{sqrt(2)=2^(1/2)}}}

{{{(2^(1/2))^(1/6)-2*root(12,2)}}} Convert from radical notation to exponential notation. So {{{root(6,2^(1/2))=(2^(1/2))^(1/6)}}}

{{{(2^(1/2))^(1/6)-2*2^(1/12)}}} Convert from radical notation to exponential notation. So {{{root(12,2)=2^(1/12)}}}

{{{2^(1/12)-2*2^(1/12)}}} Multiply the exponents {{{1/2}}} and {{{1/6}}} to get {{{1/12}}}

Take note that the GCF is {{{2^(1/12)}}}

{{{2^(1/12)(1-2)}}} Factor out the GCF {{{2^(1/12)}}}

{{{2^(1/12)(-1)}}} Subtract

{{{root(12,2)(-1)}}} Convert from exponential notation to radical notation. So {{{2^(1/12)=root(12,2)}}}

{{{-root(12,2)}}} Rearrange the terms.

So {{{root(6,sqrt(2))-root(12,2^13)=-root(12,2)}}} (Note: the right side reads the negative 12th root of 2)

Notes:

I used the following identities to simplify the expression:

1) {{{x^(y+z)=x^y*x^z}}}

2) {{{sqrt(x*y)=sqrt(x)*sqrt(y)}}}

3) {{{root(n,x^n)=x}}}```