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