Question Number 152323 by mathdanisur last updated on 27/Aug/21
$$\mathrm{x}^{\mathrm{2}} \centerdot\mathrm{y}=\frac{\mathrm{1}}{\mathrm{18}}\:\:\mathrm{and}\:\:\mathrm{x}\centerdot\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\mathrm{find}\:\:\left(\mathrm{xy}\right)^{−\mathrm{2}} \:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 27/Aug/21
$${x}^{\mathrm{2}} {y}\:=\:\frac{\mathrm{1}}{\mathrm{18}}\:\mathrm{and}\:{xy}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\:\left({x}^{\mathrm{2}} {y}\right)\left({xy}^{\mathrm{2}} \right)\:=\:\left({xy}\right)^{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{12}×\mathrm{18}}\:=\:\frac{\mathrm{1}}{\mathrm{216}} \\ $$$${xy}\:=\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{216}}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left({xy}\right)^{−\mathrm{2}} \:=\:\mathrm{6}^{\mathrm{2}} \:=\:\mathrm{36} \\ $$
Commented by mathdanisur last updated on 27/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$
Commented by otchereabdullai@gmail.com last updated on 27/Aug/21
$$\mathrm{nice}\:\mathrm{one}! \\ $$