The prob is: $2\sqrt[3]{8x^2}\,+\,5\sqrt[3]{27x^2}\,-\,3\sqrt{x^3}$

That simplifies to $4\sqrt[3]{x^2} + 15\sqrt[3]{x^2}\,-\,3x\sqrt{x}\;=\;\boxed{19\sqrt[3]{x^2}\,-\,3x\sqrt{x}}$

Lookin' good?

2. Originally Posted by Jonboy

The prob is: $2\sqrt[3]{8x^2}\,+\,5\sqrt[3]{27x^2}\,-\,3\sqrt{x^3}$

That simplifies to $4\sqrt[3]{x^2} + 15\sqrt[3]{x^2}\,-\,3x\sqrt{x}\;=\;\boxed{19\sqrt[3]{x^2}\,-\,3x\sqrt{x}}$

Lookin' good?
looks fine. still don't like that the x's in the last term are in separate places, but in some sense it is simpler i guess

3. ## Re: #2 Simplifying Radicals

You are right , one can go a step further write the last term as square-root of x^3 for similarity.