YouTube - Precalculus: The Essentials that Students Seem to Forget
It starts at 5:29. Basically, what he is saying is that:

$\displaystyle x^{\frac{1}{3}} + x^{\frac{1}{2}} - 7 = 0$

He's says that you can take the LCM of the fractional exponents (6), and multiply both sides of the equation by x^6 to remove the fractions. But, if you do that, wouldn't this happen:

$\displaystyle x^{\frac{6}{1}} * x^{\frac{1}{3}} = x^{\frac{18}{3} + \frac{1}{3}} = x^{\frac{19}{3}}$

Which doesn't help at all.. And even if you tried something like this:

$\displaystyle (x^{\frac{1}{3}} + x^{\frac{1}{2}})^6 = 7^6$

Even using the binomial theorem and expanding, it would probably reintroduce more strange fractional exponents. Am I right?