How did this binomial $(x^2+3)^{-\frac 13}-\frac 23x^2(x^2+3)^{-\frac 43}$ get factored into these $(x^2+3)^{-\frac 43}(\frac 13x^2+3)$ factors?

Question

This is an exercice in PRECALCULUS mathematics for calculus 7th, Section 1.3 Question 129. And I just can't figure out how the book came up whith the answer.

I know how to factor this binomial: $$(x^2+3)^{-\frac 13}-\frac 23x^2(x^2+3)^{-\frac 43}$$ ... into these factors (it may be wrong): $$(x^2+3)^{-\frac 43} ((x^2+3)^{\frac 23}-\frac 23x^2)$$

But from there i have no idea how to continue, I'm following the basic strategy of "pulling out" the smallest shared term $(x^2+3)^{-\frac 43}$

Answer 1

HINT: As $-\frac{1}{3}=-\frac{4}{3} +1$, it follows that $(x^2+3)^{-\frac{1}{3}} = (x^2+3)^{-\frac{4}{3}}(x^2+3)$. In other words as noted in the comments already, your exponent of $\frac{2}{3}$ should be replaced by 1 as the exponent. Then continue as you would have before...

Answer 2

Since $$ (x^2+3)^{-\frac{1}{3}} = (x^2+3)^{-\frac{4}{3}}(x^2+3) $$ The original one becomes $$ (x^2+3)^{-\frac 13}-\frac 23x^2(x^2+3)^{-\frac 43}= (x^2+3)^{-\frac{4}{3}}(x^2+3)-\frac 23x^2(x^2+3)^{-\frac 43}\\= (x^2+3)^{-\frac{4}{3}}\left (\dfrac 13x^2+3\right ) $$ We are done.

Answer 3

The factorisation is wrong. The main idea is this: that when in the binomial $ab+ac,$ you factor out $a,$ then what you're in effect doing is rewriting the sum as $$a\left(\frac{ab}{a}+\frac{ac}{a}\right).$$

Hence by pulling out $(x^2+3)^{-4/3},$ you have to divide each summand by $(x^2+3)^{-4/3},$ or in other words multiply each summand by $(x^2+3)^{4/3},$ which gives you $$(x^2+3)^{-4/3}\left(x^2+3-\frac23 x^2\right),$$ from where you can continue.