How do you simplify $(3x^3y)/((3y)^-2)$ ?

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Answer 1 · 2018-05-10T19:58:22

$27x^3y^3$

$1/(3y)^(-2)=(3y)^2$

$(3x^3y)/(3y)^(-2)=3x^3y xx (3y)^2$

$=3x^3y xx 9y^2=27x^3y^3$

Answer 2 · 2018-05-10T20:02:10

$(3x^3y)/(3y)^(-2)$ = $=27x^3y^3$ = $(3xy)^3$

You want to simplify $(3x^3y)/(3y)^(-2)$

We must remember that $1/x^-n=x^n$
Therefore $1/(3y)^(-2)=(3y)^2$

We, therefore, can write:
$(3x^3y)/(3y)^(-2)$
$=(3x^3y)(3y)^2$
$=3x^3y*3^2y^2$
$=27x^3y^3$

As $27=3^3$ it's perhaps a little more elegant if we write this
$(3xy)^3$

How do you simplify (3x^3y)/((3y)^-2)(3x^3y)/((3y)^-2)?