Question 606547
Radicals are cumbersome; fractional exponents are easier to work with.
The only problem is writing those fractional exponents so that they are clearly readable.
If your problem was {{{(root(4,27))*(root(3,3))}}} I would solve it like this
{{{(root(4,27))*(root(3,3))=(root(4,3^3))*(root(3,3))=(3^(3/4))*(3^(1/3))}}}
I can write that final expressions as (3^(3/4))*(3^(1/3)) and hope it is more readable.
That is the product of two powers of base 3, which equals a power of base 3 whose exponent is the sum of the exponents.
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)
I can add fractions:
{{{3/4+1/3=9/12+4/12=(9+4)/12=13/12}}} so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(13/12)={{{root(12,3^13)}}}
Or even better,
{{{root(12,3^13)=root(12,(3^12*3))=root(12,3^12)*(root(12,3))}}}={{{3}}}{{{root(12,3)}}}
We could have kept writing fractional exponents to the end because
{{{3/4+1/3=13/12=1+1/12}}} so
(3^(3/4))*(3^(1/3))=3^(3/4+1/3)=3^(1+1/12)=(3^1)*(3^(1/12))={{{3}}}{{{root(12,3)}}}
(I would like to write that front 3 a little lower, as I would with pen and paper, but I can't. I hope you understand what I mean, anyway).