Does anyone understand what this question is asking?

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Does anyone understand what this question is asking?

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Answer 1 · 2017-01-20T11:51:58

$= x \frac{3^{\frac{2}{3}}}{3}$ $=x3^(2/3)/3$ .

Explanation:

The expression is

$\frac{x}{3^{\frac{1}{3}}}$ $x/3^(1/3)$

Multiply and divide by the same number #3^(2/3)

$= x 3^{\frac{2}{3}} \frac{1}{3^{\frac{2}{3}} (3^{\frac{1}{3}})}$ $=x3^(2/3)1/(3^(2/3)(3^(1/3))$

Now, for the denominator, use $a^{m} a^{n} = a^{m + n}$ $a^m a^n=a^(m+n)$ .

$= x \frac{3^{\frac{2}{3}}}{3^{\frac{2}{3} + \frac{1}{3}}}$ $=x3^(2/3)/(3^(2/3+1/3))$

$= s \frac{3^{\frac{2}{3}}}{3^{1}}$ $=s3^(2/3)/3^1$

$= x \frac{3^{\frac{2}{3}}}{3}$ $=x3^(2/3)/3$ .

Use, $a^{m n} = {(a^{m})}^{n}$ $a^(mn)=(a^m)^n$ r

It becomes

$x \frac{{(3^{2})}^{\frac{1}{3}}}{3}$ $x(3^2)^(1/3)/3$

$= x \frac{9^{\frac{1}{3}}}{3}$ $=x9^(1/3)/3$

Thanks to Andrea for identifying this as the print in ( D ).

Answer 2 · 2017-01-20T18:27:16

Here is a longer answer to your question. (Using different notation.)

Explanation:

We start with $\frac{x}{\sqrt[3]{3}} =$ $x/root3(3) =$

Looking at the list of choices, we see that each choice has an $x$ $x$ in it. So we are not "evaluating at " some number for $x$ $x$ . But we do see an equal sign, so the answer must be equal to the starting expression.

Looking at the starting expression and the choices of answers, we see that one difference is that the start has a radical (a root) in the denominator and none of the answer choices have that.

What the question is asking us to do is to write an expression that is equivalent to the start, but has no radical in the denominator.
(It's like writing an equal number, but "expression" because of the variable).

The number $\sqrt[3]{3}$ $root3(3)$ in the denominator is irrational but the choices have rational denominators.

AHA! We are being asked to rationalize a denominator. ("Rationalize" mean "make rational".)

Now, for square roots, we know that $\sqrt{a^{2}} = a$ $sqrt(a^2) = a$ for any number $a$ $a$ . But this question involves a $3^{rd}$ $3^"rd"$ root.

The third root has the property that $\sqrt[3]{a^{3}} = a$ $root3(a^3) = a$

So we would like to see $\sqrt[3]{3^{3}}$ $root(3)(3^3)$ in the denominator.

We already have $\sqrt[3]{3}$ $root3(3)$ , so we need 2 more $3$ $3$ 's under the radical.

We'll multiply $\frac{x}{\sqrt[3]{3}}$ $x/root3(3)$ by $1$ $1$ in the form $\frac{\sqrt[3]{3^{2}}}{\sqrt[3]{3^{2}}}$ $root3(3^2)/root3(3^2)$

$\frac{x}{\sqrt[3]{3}} = \frac{x}{\sqrt[3]{3}} \cdot \frac{\sqrt[3]{3^{2}}}{\sqrt[3]{3^{2}}}$ $x/root3(3) = x/root3(3) * root3(3^2)/root3(3^2)$

$= \frac{x \sqrt[3]{3^{2}}}{\sqrt[3]{3} \sqrt[3]{3^{2}}}$ $= (xroot3(3^2))/(root3(3)root3(3^2))$

$= \frac{x \sqrt[3]{3^{2}}}{\sqrt[3]{3^{3}}}$ $= (xroot3(3^2))/root3(3^3)$

$= \frac{x \sqrt[3]{3^{2}}}{3}$ $= (xroot3(3^2))/3$

This is not one of the choices, but $3^{2} = 9$ $3^2 = 9$ , so we can see that pur answer is equal to $D$ $D$ .

Note

I think the solution would be clearer if we had $\frac{x}{\sqrt[3]{5}}$ $x/root3(5)$ , because the two different occurnces of the numeral $3$ $3$ might be confusing.

$\frac{x}{\sqrt[3]{5}} = \frac{x}{\sqrt[3]{5}} \cdot \frac{\sqrt[3]{5^{2}}}{\sqrt[3]{5^{2}}}$ $x/root3(5) = x/root3(5) * root3(5^2)/root3(5^2)$

$= \frac{x \sqrt[3]{5^{2}}}{\sqrt[3]{5} \sqrt[3]{5^{2}}}$ $= (xroot3(5^2))/(root3(5)root3(5^2))$

$= \frac{x \sqrt[3]{5^{2}}}{\sqrt[3]{5^{3}}}$ $= (xroot3(5^2))/root3(5^3)$

$= \frac{x \sqrt[3]{5^{2}}}{5}$ $= (xroot3(5^2))/5$

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