How to simplify a polynomial with fractional exponents

Question

I am trying to simplify the following polynomial with fractional exponents.

I have $3x^\frac{5}{3}-\frac{5x^\frac{2}{3}}{3}-\frac{4x^\frac{-1}{3}}{3}$. How do I simplify it so that I get $\frac{(x-1)(9x+4)}{3x^\frac{1}{3}}$?

I can simplify to $\frac{9x^\frac{5}{3}-5x^\frac{2}{3}-4}{3x^\frac{1}{3}}$, but don't know how to simplify any further.

Answer 1

$$3x^\frac{5}{3}-\frac{5x^\frac{2}{3}}{3}-\frac{4x^\frac{-1}{3}}{3}=\dfrac {9x^{\frac 5 3}-5x^{\frac 2 3}-4x^{\frac {-1} 3}}{3}$$

Now, you can multiply both the numerator and the denominator of $\dfrac {9x^{\frac 5 3}-5x^{\frac 2 3}-4x^{\frac {-1} 3}}{3}$ by $x^{\frac 1 3}$

$\dfrac {\left ( 9x^{\frac 5 3}-5x^{\frac 2 3}-4x^{\frac {-1} 3}\right )\cdot x^{\frac 1 3}}{3\cdot x^{\frac 1 3}}= \dfrac { 9x^{\frac 5 3+ \frac 1 3}-5x^{\frac 2 3 + \frac 1 3}-4x^{\frac {-1} 3+ \frac 1 3}}{3x^{\frac 1 3}}=\dfrac { 9x^{2}-5x-4}{3x^{\frac 1 3}}$

Answer 2

So I figured out how to do it. I had to pull out $x^\frac{-1}{3}$ in the numerator, so then I get $\frac{(x^\frac{-1}{3})(9x^2-5x-4)}{3}$, which results in

$\frac{9x^2-9x+4x-4}{3x^\frac{1}{3}}$ and this simplifies to $\frac{(x-1)(9x+4)}{3x^\frac{1}{3}}$

Answer 3

When you simplify, you wrongly pull out $x^{1/3}$ - a trivial mistake on the 4th-grade level.