(4xy^2)^-3/7?

(4xy^2)^-3/7?

\((4xy^2)^{-3}/7\\ =(4xy^2)^{-3}\div7\\ =\frac{1}{(4xy^2)^{3}}\div7\\ =\frac{1}{(64x^3y^6)}\times \frac{1}{7}\\ =\frac{1}{448x^3y^6}\\\)

OR

\((4xy^2)^{-3/7}\\ =\frac{1}{(4xy^2)^{3/7}}\\ =\frac{1}{(64x^3y^6)^{1/7}}\\ =\frac{1}{\sqrt[7]{64x^3y^6}}\\\)

The first one is the 'correct' interpretation, the second one needed brackets added. :)

Sorry guest, I only just realised that yours its the same as mine    

Simplify the following:
1/(7 (4 x y^2)^3)

Multiply each exponent in 4 x y^2 by 3:
1/(7×4^3 x^3 y^(3×2))

3×2  =  6:
1/(4^3×7 x^3 y^6)

4^3 = 4×4^2:
1/(4×4^2×7 x^3 y^6)

4^2 = 16:
1/(4×16×7 x^3 y^6)

4×16  =  64:
1/(64×7 x^3 y^6)

64×7  =  448:
Answer: | 1/(448 x^3 y^6)