{(a^-3/m)(b^6/n) / (a^-6/m)(b^9/n)}^ -1/3?

Question

simplify the expression. Assume (a) and (b) are positive real numbers and (m) and (n) are rationalal numbers.

Answer 1

{(a^-3/m)(b^6/n) / (a^-6/m)(b^9/n)}^ -1/3
= {(a^(-3)/m)*(b^6/n) * (m/(a^(-6)) * (n/(b^9))} ^(-1/3)
= {(a^(-3)) * (b^6)*(1/(a^(-6))*(1/(b^9))} ^(-1/3)
= {(1/(a^(-3))*(1/(b^3))}^(-1/3)
= {a^3*b^(-3)}^(-1/3)
= {a^(-1)*b}
= b/a
-

Answer 2

{(a^-3/m)(b^6/n) / (a^-6/m)(b^9/n)}^ -1/3? \

{(a^-3/m)(b^6/n) * (m/a^-5)(n/b^9 )}^-1/3 =

= {(a^-3 * b^6)/mn * mn/(a^-5 * b^9)}^-1/3 =

= {(a^-3 * b^6)/(a^-5 * b^9)}^-1/3 =

= {a^-3 * b^6 * a^5 * b^-9} ^ -1/3 =

= (a^2 * b^-3)^-1/3 = (a^2/b^3)^-1/3 =

= 1/(a^2/b^3)^(1/3) = (b^3/a^2)^1/3 =

= b / a^(2/3) = [b*a^(1/3)]/a