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So I feel totally stupid asking this considering I am in precalc but our professor threw this question at us. I have the answer, but he didn't provide the process:

Rationalize the denominator:


The answer is:


Please explain it how he got there. I know you have to multiply by


But I end up with: $$\frac{\sqrt[7]{x^4}}{x^4}$$

What am I doing wrong?

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What if you write it as: $\dfrac{x^{3/7}}{x^1} = \dfrac{x^{3/7}}{x^{7/7}}$? Does that help? – Amzoti Dec 7 '13 at 0:55
up vote 0 down vote accepted

You have to multiply it by $\frac {\sqrt [7]{x^3}}{\sqrt [7]{x^3}}$ The numerator in the answer shows that. Then ${\sqrt [7]{x^3}}\cdot {\sqrt [7]{x^4}}={\sqrt [7]{x^7}}=1$. I find it easier to use fractional exponents, changing ${\sqrt [7]{x^4}}$ to $x^{\frac 47}$ and proceeding. You could do that and see if it helps.

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It's coming back to me now. I need to fill in the fraction basically. – munchschair Dec 7 '13 at 1:10

What you're doing wrong is that you're following a rule that says you should multiply the numerator and denominator by $\sqrt[7]{x}$.

One the bottom you've got $$ \sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}. $$ What you need on the bottom to get rid of the radical is $$ \sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}. $$ Multiply that out an you just have $x$, with no radical.

So you have to multiply both the numerator and the denominator by $\sqrt[7]{x}^3$.

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