Whenever you encounter a quantitative question with answer choices, be sure to SCAN the answer choices before performing any calculations. In many cases, the answer choices provide important clues regarding how to best solve the question.
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Comment on Quotient with Cube Roots
I converted the cubed roots
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
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