How to Simplify a Tricky Fraction?

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To simplify the fraction \(\frac{\sqrt[3]{m^2+m} \cdot \sqrt{1+m^2}}{\sqrt{1+m^2} \cdot \sqrt{1+m^2}}\), the denominator simplifies to \(1+m^2\). The numerator remains as \((m^2+m)^{\frac{1}{3}} \cdot (1+m^2)^{\frac{1}{2}}\). Since the two factors in the numerator have different bases, they cannot be combined further. The only simplification possible is to cancel one \(\sqrt{1+m^2}\) from the numerator and denominator, resulting in \(\frac{(m^2+m)^{\frac{1}{3}}}{\sqrt{1+m^2}}\). This is the most simplified form of the expression.
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Homework Statement



How do I simplify the following

\frac{\sqrt[3]{m^2+m} . \sqrt{1+m^2}}{\sqrt{1+m^2}.\sqrt{1+m^2}}


The Attempt at a Solution



I know that the denominator will be 1+m^2 but I don't know how to simplify the numerator. Can anyone show me how?
 
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Try writing out the powers as a fraction. That is, let the square roots be ^(1/2) and such. It'll be easier to see as you just add the fractions.
 
Anonymous217 said:
Try writing out the powers as a fraction. That is, let the square roots be ^(1/2) and such. It'll be easier to see as you just add the fractions.

It doesn't seem to change much

\frac{(m^2+m)^{\frac{1}{3}} . (1+m^2)^{\frac{1}{2}}}{1+m^2}

Should we just add the powers, and what about the terms?
 
Last edited:
What's 1/3 + 1/2?
 
Anonymous217 said:
What's 1/3 + 1/2?
I don't see how that is applicable in this problem. If I'm understanding you correctly, you are trying to convince us that a1/3b1/2 can somehow be combined.
 
math2010 said:
It doesn't seem to change much

\frac{(m^2+m)^{\frac{1}{3}} . (1+m^2)^{\frac{1}{2}}}{1+m^2}

Should we just add the powers, and what about the terms?
I think this is about all you can do by way of simplification. The two factors in the numerator have different bases, so can't be combined.
 
Are you sure the first term is m^2+m not m^3+m? cus then, you could factor out an m and go from there.

Otherwise, I see no way of simplifying this expression other than canceling out one sqrt(1+m^2) from top and bottom.
 
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