Answer:
c = 6 and d = 2
Step-by-step explanation:
As the given equation is
[tex]\sqrt[3]{162x^cy^5}=3x^2y\left(\sqrt[3]{6y^d}\right)[/tex]
Taking cube on both sides,
[tex]\left(\sqrt[3]{162x^cy^5}\right)^3=\left(3x^2y\left(\sqrt[3]{6y^d}\right)\right)\:^3[/tex]
[tex]162x^cy^5=(3x^2y)^3(\sqrt[3]{6y^d})^3\\\\162x^cy^5=27x^{2\times3}y^3(6y^d)....[(a^m)^n=a^{mn}]\\\\162x^cy^5=162x^6y^{3+d}....[a^{m}\times\ a^n=a^{m+n}]\\\\\text{Comparing the power of corresponding variables}\\\\\ c=6\ ;\ 3+d=5\Rightarrow\ d=5-3=2[/tex]
So, c = 6 and d = 2
