How do you simplify #((3x)/(4y))^3 -: (81x^2y^-6)^-(1/4)#?

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Answer 1 · 2017-02-17T11:10:17

#(3^4x^(7/2))/(4^3y^(9/2)#

#((3x)/(4y))^3-:(81x^2y^-6)^-(1/4)#

#:.(3^3x^3)/(4^3y^3)-:(3^(-4/4)*x^(-2/4)*y^(6/4))#

#:.(3^3x^3)/(4^3y^3)-:(3^-1*x^(-1/2)*y^(3/2))#

#:.(3^3x^3)/(4^3y^3) xx 1/(3^-1*x^(-1/2)*y^(3/2))" "# (multiply by the reciprocal)

#:.(3^(3-(-1))*x^(3-(-1/2)))/(4^3*y^(3+3/2)#

#:.(3^4*(x^(7/2)))/(4^3*y^(9/2))#

#:.(3^4x^(7/2))/(4^3y^(9/2)#