how to simplify a function with fractional exponents

Question

I want to understand how to simplify fractional exponents in the formula $$g(x) = (4/3)x^{-1/3}-(5/3)x^{2/3}$$

My text book says the answer is $$\frac{4 -5x}{3x^{1/3}}$$

but I can only simplify it to $$\frac{4x^{-1/3}-5x^{2/3}}{3}$$

Appreciate any help

Answer 1

Recall that $x^{-1/3} = \frac{1}{x^{1/3}}$. Using this rule, we can simplify to $$g(x)= \frac{4}{3x^{1/3}} - \frac{5x^{2/3}}{3}$$

Now need a common denominator to subtract these two fractions. If we multiply the second fraction by $\frac{x^{1/3}}{x^{1/3}}$, we get $$g(x)= \frac{4}{3x^{1/3}} - \frac{5x^{2/3}}{3} \cdot \frac{x^{1/3}}{x^{1/3}}= \frac{4}{3x^{1/3}} - \frac{5x^{2/3+1/3}}{3x^{1/3}}=\frac{4}{3x^{1/3}} - \frac{5x}{3x^{1/3}} = \frac{4-5x}{3x^{1/3}}$$

Answer 2

Simply factor out $x^{-1/3}$: $$ \frac 4 3x^{-1/3}-\frac53x^{2/3}=\frac{x^{-1/3}}3\bigl(4-5x^{2/3-(-1/3)}\bigr)=\frac1{3x^{1/3}}(4-5x^{3/3}).$$

Answer 3

Since all the exponents involve "thirds", you could use a change of variable, $u=x^{1/3}$, and rewrite $$ \begin{aligned} \frac{4x^{-1/3}-5x^{2/3}}{3} =& \frac{4u^{-1}-5u^2}{3} \\ =& \cfrac{\cfrac{4}{u}-5u^2}{3} \\ =& \frac{4-5u^3}{3u} \end{aligned} $$ Then substitute back $u=x^{1/3}, u^3=x$, to get the desired answer

Answer 4

They multiplied the top and the botumn of your answer by $x^{1/3}$ to look a little bit nicer.

$$ \frac{4x^{-1/3}-5x^{2/3}}{3}=\frac{x^{1/3}(4x^{-1/3}-5x^{2/3})}{x^{1/3}(3)}=$$

$$ \frac{4 -5x}{3x^{1/3}}$$