Question

Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?

Answer 1

Answer:- a.The given expression is equivalent to  $\frac{x^{10}y^{14}}{729}$

Given expression:- $[\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}$

$=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}$

$=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}$

$=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}$

$=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}$

$=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}$

Thus a. is the right answer.

Answer 2

Answer:

a). $=\frac{x^{10}*y^{14}}{729}$

Step-by-step explanation:

$[(\frac{(3*x*y^{5})^{3}}{(x^{-2}*y^{2})^{-4}})]^{-2}$

Taking the first factor development the operation, knowing first resolve the inside operation

$=(3*x*y^{-5})^{3}\\=(3^{3}*x^{3}*y^{3*5})\\=(27*x^{3}*y^{-15})$

Do the same with the other expression

$=(x^{-2}*y^{2})^{-4}\\=(x^{-2*-4}*y^{2*-4})\\=(x^{8}*y^{-8})$

So the expression is now:

$=[\frac{27*x^{3}*y^{-15}}{x^{8}*y^{-8}}]^{-2}\\Up \\=27^{-2}*x^{3*-2}*y^{-15*-2}\\=-729*x^{-6}*y^{30} \\Down\\=x^{8*-2}*y^{-8*-2}\\==x^{-16}*y^{16}$

$=\frac{-729*x^{-6}*y^{30}}{x^{-16}*y^{16}}\\=-729*x^{-6}*x^{16}*y^{30}y^{-14}\\=-729*x^{10}*y^{16}\\=729^{-1}*x^{10}*y^{16}\\=\frac{x^{10}*y^{16}}{729}$