Question 918849
{{{sqrt(20*(root(3, 135)))}}} ?
 
The factorization of each number will tell you if you can simplify.
{{{135=27*5=3*3*3*5=3^3*5}}} and {{{20=4*5=2*2*5=2^2*5}}} so there is room for some simplifications.
What is the simplest (or more convenient) way to express it?
Would you like to show your work with roots, or with fractional exponents?
 
WITH ROOTS:
{{{sqrt(20*(root(3, 135)))}}}={{{sqrt(4*5*(root(3,27*5)))}}}={{{sqrt(4)*sqrt(5)*sqrt((root(3,27)*root(3,5)))}}}={{{2*sqrt(5)*sqrt((3*root(3,5)))}}}={{{2*sqrt(5)*sqrt(3)*sqrt(root(3,5))}}}={{{2*sqrt(5)*sqrt(3)*root(6,5)}}}={{{2*sqrt(3)*sqrt(5)*root(6,5)}}}={{{2*sqrt(3)*root(6,5^3)*root(6,5)}}}={{{2*sqrt(3)*root(6,5^3*5)}}}={{{2*sqrt(3)*root(6,5^4)}}}={{{2*sqrt(3)*root(3,5^2)}}}={{{2*sqrt(3)*root(3,25)}}}
 
WITH FRACTIONAL EXPONENTS:
{{{sqrt(20*(root(3, 135)))}}}={{{(20*135^(1/3))^(1/2)}}}={{{(20)^(1/2)*(135^(1/3))^(1/2)}}}={{{(2^2*5)^(1/2)*((3^3*5)^(1/3))^(1/2)}}} ={{{(2^2)^(1/2)*5^(1/2)*((3^3)^(1/3)*5^(1/3))^(1/2)}}}=
{{{2*5^(1/2)*(3*5^(1/3))^(1/2))}}}={{{2*5^"(1/2)"*3^"(1/2)"*5^("(1/3)"*"(1/2)")}}}={{{2*5^"(1/2)"*3^"(1/2)"*5^"(1/6)"}}}={{{2*3^"(1/2)"*5^(("1/2"+"1/6"))}}}={{{2*3^"(1/2)"*5^"(4/6)"}}}={{{2*3^"(1/2)"*5^"(2/3)"}}}