Question

What values of c and d make the equation true? RootIndex 3 StartRoot 162 x Superscript c Baseline y Superscript 5 Baseline EndRoot = 3 x squared y (RootIndex 3 StartRoot 6 y Superscript d Baseline EndRoot) c = 2, d = 2 c = 2, d = 4 c = 6, d = 2 c = 6, d = 4

Answer 1

Answer:

its c

Step-by-step explanation:

Answer 2

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

$\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}$

Let's cube both sides of the equation:

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

The left side just simplifies the cubic root with the cube:

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

On the right side, we'll simplify the cubic root where possible and power what's outside of the root:

$162x^cy^5=3^3x^6y^3 (6y^d)$

Simplifying

$x^cy^5=x^6y^{3+d}$

Equating the powers of x and y separately we find

c=6

5=3+d

d=2

The values are

$\boxed{c=6,d=2}$