What values of c and d make the equation true? mc011-1.jpg c = 2, d = 2 c = 2, d = 4 c = 6, d = 2 c = 6, d = 4

If the equation is ∛(162x^cy^5) = 3x²y(∛(6y^d)
We can simplify the equation by first cubing both sides
            = 162x^cy^5 = (27x^6y^3)6y^d
            = 162x^cy^5 = 162 x^6y^3y^d
dividing both sides by 162
           = x^cy^5 =x^6y^3y^d
      equating the values with similar bases
   x^c = x^6, therefore c=6
and y^5=y^3y^d
        y ^5 = y^(3+d)
hence, 5 = 3+d
thus, d= 2
Therefore, the values of d and c that will satisfy the equation are, c =6 and d=2