What values of c and d make the equation true? Assume and

c = 1, d = 3

c = 1, d = 32

c = 2, d = 8

c = 2, d = 32

The equation for this problem is

∛ (162 x^c y⁵) = 3x²y ∛ (6y^d)

First, we take the cube of the two sides of the equation

162 x^c y⁵ = 27 x⁶ y³ (6 y^d)

Simplifying

x^c y⁵ = x⁶ y^ (3 + d)

x^c

Next, equate the exponents

For x:

c = 2

For y:

5 = - 3 + d

d = 8

The answer is

c = 2 and d = 8