Question

What values of c and d make the equation true? sqrt[3] 162x^ 6 y^ 5 =3x^ 2 y sqrt[3] 6y^ d c = 2 , d = 2 c = 2 d = 4 c = 6 d = 2 c = 6 d = 4

Answer 1

Answer: c=6 and d=2

Step-by-step explanation:

Answer 2

c = 6 and d = 2

Explanation:

As the given equation is

`\sqrt[3]{162x^cy^5}=3x^2y\left(\sqrt[3]{6y^d}\right)`

Taking cube on both sides,

`\left(\sqrt[3]{162x^cy^5}\right)^3=\left(3x^2y\left(\sqrt[3]{6y^d}\right)\right)\:^3`

`162x^cy^5=(3x^2y)^3(\sqrt[3]{6y^d})^3\\\\162x^cy^5=27x^(2*3)y^3(6y^d)....[(a^m)^n=a^(mn)]\\\\162x^cy^5=162x^6y^(3+d)....[a^(m)*\ a^n=a^(m+n)]\\\\\text{Comparing the power of corresponding variables}\\\\\ c=6\ ;\ 3+d=5\Rightarrow\ d=5-3=2`

So, c = 6 and d = 2