(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)
(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