Answer:
c=6, d=2
Step-by-step explanation:
Equations
We must find the values of c and d that make the below equation be true
[tex]\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}[/tex]
Let's cube both sides of the equation:
[tex]\left (\sqrt[3]{162x^cy^5}\right )^3=\left (3x^2y \sqrt[3]{6y^d}\right)^3[/tex]
The left side just simplifies the cubic root with the cube:
[tex]162x^cy^5=\left (3x^2y \sqrt[3]{6y^d}\right)^3[/tex]
On the right side, we'll simplify the cubic root where possible and power what's outside of the root:
[tex]162x^cy^5=3^3x^6y^3 (6y^d)[/tex]
Simplifying
[tex]x^cy^5=x^6y^{3+d}[/tex]
Equating the powers of x and y separately we find
c=6
5=3+d
d=2
The values are
[tex]\boxed{c=6,d=2}[/tex]
