# Question: Quotient with Cube Roots

## Comment on Quotient with Cube Roots

### I converted the cubed roots

I converted the cubed roots to powers. I got (4)((6^(2/3))/(8)((6^1/3)) and got A. What did I do wrong? ### You almost had it. You forgot

You almost had it. You forgot to raise 4 to the power of 2.

RULE: (xy)^k = (x^k)(y^k)

So, for the part inside the brackets in the numerator, we can first combine the 4 roots to get 4[6^(1/3)]

At this point we must raise 4[6^(1/3)] to the power of 2. So, by the above RULE, we get: 4² times [6^(1/3)]²

So, the numerator simplifies to be (16)(6^(2/3))

### You have multiplied the

You have multiplied the denominator values of inside the roots, but why u didn't multiply the values of upside .. four times 6 inside the roots, please reply me i bought the video's because I have to give gre soon. ### I think you are mistaking

I think you are mistaking multiplying roots for multiplying algebraic expressions.

For example, 3(2x + 5) = 6x + 15. Here, we are multiplying the 3 by EACH expression in the brackets. Notice that the brackets contain a SUM.

Likewise, 5(4x - 11) = 20x - 55. Here, we are multiplying the 5 by EACH expression in the brackets. Notice that the brackets contain a DIFFERENCE.

If we are multiplying a PRODUCT be some value, we don't perform the same operations as we did with we're multiplying a SUM or DIFFERENCE

For example, (3)(2x) = 6x and NOT (3)(2)(3)(x)
The reason for this is that (3)(2x) = (3)(2)(x) = 6x

Likewise, (2√2)(4√3) = (2)(√2)(4)(√3)
= (2)(4)(√2)(√3)
= 8√6

For more on this, watch https://www.greenlighttestprep.com/module/gre-powers-and-roots/video/1042