Question 310502
We're going to use the following identities 


1) {{{(x^(y))/(x^(z))=x^(y-z)}}}



2) {{{(x^(y))^z=x^(y*z)}}}



3) {{{(a/b)^x=(a^x)/(b^x)}}}



{{{((27b^(1/3)c^(3/4)^"")/(8b^(-2/3)c^(1/2)^""))^(4/3)^""}}} Start with the given expression.



{{{((27b^(1/3-(-2/3))c^(3/4-1/2)^"")/8)^(4/3)^""}}} Use identity # 1 to divide the variable terms. 



{{{((27b^(1/3+2/3)c^(3/4-1/2)^"")/8)^(4/3)^""}}} Rewrite {{{1/3-(-2/3)}}} as {{{1/3+2/3}}}



{{{((27b^(1/3+2/3)c^(3/4-2/4)^"")/8)^(4/3)^""}}} Multiply {{{1/2}}} by {{{2/2}}} to get {{{2/4}}}



{{{((27b^((1+2)/3)c^((3-2)/4)^"")/8)^(4/3)^""}}} Combine the fractions by combining the numerators over the common denominator.



{{{((27b^(3/3)c^(1/4)^"")/8)^(4/3)^""}}} Combine like terms.



{{{((27b^1c^(1/4)^"")/8)^(4/3)^""}}} Reduce.



{{{((27b^1c^(1/4)^"")^(4/3)^"")/(8^(4/3)^"")}}} Now use identity # 3



{{{((27b^1c^(1/4)^"")^(4/3)^"")/((2^3)^(4/3)^"")}}} Rewrite {{{8}}} as {{{2^3}}}



{{{((27b^1c^(1/4)^"")^(4/3)^"")/((2^(3(4/3))^""))}}} Use identity # 2 to multiply the exponents.



{{{((27b^1c^(1/4)^"")^(4/3)^"")/((2^(12/3)^""))}}} Multiply 3 and {{{4/3}}} to get {{{12/3}}}



{{{((27b^1c^(1/4)^"")^(4/3)^"")/(2^4)}}} Reduce.



{{{((27b^1c^(1/4)^"")^(4/3)^"")/(16)}}} Evaluate {{{2^4}}} to get 16



{{{((3^3b^1c^(1/4)^"")^(4/3)^"")/(16)}}} Rewrite 27 as {{{3^3}}}



{{{(3^(3(4/3))b^(1(4/3))c^((1/4)(4/3))^"")/(16)}}} Use identity # 2 to multiply the exponents.



{{{(3^(12/3)b^(4/3)c^(4/12)^"")/(16)}}} Multiply 3 and {{{4/3}}} to get {{{12/3}}}



{{{(3^4b^(4/3)c^(1/3)^"")/(16)}}} Reduce.



{{{(81b^(4/3)c^(1/3)^"")/(16)}}} Raise 3 to the 4th power to get 81




So {{{((27b^(1/3)c^(3/4)^"")/(8b^(-2/3)c^(1/2)^""))^(4/3)^""=(81b^(4/3)c^(1/3)^"")/(16)}}} where every variable is positive.