Simple equation simplification

Question

Looking tp get some clarification on the following simplifying of a equation.

how do you go from $$x=r(rx-x^3)-(rx-x^3)^3$$

to

$$x(x^2-r+1)(x^2-r-1)(x^4-rx^2+1)=0$$

Answer 1

Clearly $x$ is a factor.

Removing $x$ gives $$1=r\color{black}{(r-x^2)}-x^2\color{black}{(r-x^2)}^3\qquad \cdots (1)$$

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First method

Substituting $\color{black}{(r-x^2)}=\mp 1$gives a consistent result, so $(x^2-r\pm 1)$ are also factors.

That leaves the last factor which can be worked out easily (or not) by equating coefficients. Alternatively, substituting $x^2=\frac 1{r-x^2}$ in ($1$) gives $1=x^2(r-x^2)$ which is consistent with the substitution. Hence $x^2(r-x^2)-1$ is a factor. Multiplying by $-1$ and expanding gives the last factor in a different form as $x^4-rx^2+1$.

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Second method

Putting $u=r-x^2$ (i.e. $r=u+x^2$) in ($1$) gives $$\begin{align} 1&=ru-x^2u^3\\ x^2u^3-ru+1&=0\\ x^2u^3-u^2-x^2u+1&=0\\ (x^2u-1)(u^2-1)&=0\\ (x^2u-1)(u-1)(u+1)&=0\\ (x^2(r-x^2)-1)(r-x^2-1)(r-x^2+1)&=0\\ (-x^4+rx^2-1)(r-x^2-1)(r-x^2+1)&=0\\ (x^4-rx^2+1)(x^2-r+1)(x^2-r-1)&=0 \end{align}$$

Multiplying throughout by $x$ gives the factorisation required.

Answer 2

HINT: use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ sorry I misread this we have $$r(r-x^2)-x^2(r-x^2)^3=(r-x^2)(r-x^2(r-x^2)^2)=- \left( -{x}^{2}+r \right) \left( {x}^{6}-2\,r{x}^{4}+{r}^{2}{x}^{2} -r \right) $$ sorry for my mistake! $$rx-x^3=x(r-x^2)$$ $$\left(x(r-x^2)\right)^3=x^3(r-x^2)^3$$