# Solve $(x^3+1)=2\sqrt[3]{2x-1}$ algebraically?

I'm trying to solve the said equation in the thread title algebraically.

$$(x^3+1)=2\sqrt[3]{2x-1}$$ Cubing both sides and simplifying:
$$x^9+3x^6+3x^3-16x+9 = 0$$

Not sure if this can be solved algebraically?

Edit: WA gives $$3$$ solutions $$x=1,\frac{1}{2}(-1-\sqrt{5}),\frac{1}{2}(-1+\sqrt{5})$$

• Can you factor out $(x-1)$ or find other rational roots? Commented Aug 17, 2022 at 3:20
• @TymaGaidash $(x-1)(x^8 + x^7 + x^6 + 4 x^5 + 4 x^4 + 4 x^3 + 7 x^2 + 7 x - 9)$. It's still 8-degree polynomial. Commented Aug 17, 2022 at 3:22
• Is there a reason to expect nice solutions? Where does the problem come from?
– anon
Commented Aug 17, 2022 at 3:24
• It's interesting to observe after the fact that if $\frac 12(-1\pm \sqrt 5)$ are roots then $(x+1-\sqrt 5)(x+1 +\sqrt 5) = (x+1)^2 - 5$ so $(x-1)((x+1)^2-5)$ factors out What's left is presumably an irreducible $6$ degree. ... also interesting $x^8 +x^7+x^6 +4x^5 + 4x^4 + 4x^3 +7x^2 + 7x + 7 =16$ factors to $(x^2+x+1)(x^6+4x^3+ 7)=16$ which is weird. Not sure any of that helps. Commented Aug 17, 2022 at 3:57
• This is the coffin problem number 12 tanyakhovanova.com/Coffins/coffinsmain.html
– user561334
Commented Aug 17, 2022 at 4:09

Rearrange as $$\underbrace{\frac{x^3+1}{2}}_{f(x)}=\underbrace{\sqrt[3]{2x-1}}_{g(x)}$$ Since $$f(x)$$ is a bijective function on $$\mathbb R$$, it must have an inverse. But note that the inverse of $$f(x)$$ is $$g(x)$$. Thus, if the two curves intersect, they must intersect ON the line $$y=x$$ (because $$f$$ and $$g$$ are mirror images about this line).

Thus we solve the two curves $$y=\dfrac{1+x^3}{2}$$ and $$y=x$$ to get $$x^3-2x+1=0$$ which boils down to $$(x-1)(x^2+x-1)=0$$ which can be readily solved to get solutions $$\displaystyle 1, \frac{-1\pm\sqrt5}{2}$$.

• Great observation! +1 Commented Aug 17, 2022 at 3:42
• @KLM_Cubed make sure to put the solutions in the original equation to verify that they all satisfy the original. Commented Aug 17, 2022 at 3:45
• "if the two curves intersect, they must intersect ON the line y=x" $\;$ No, why? At least not without additional assumptions, see for example Roots of $f(x)=f^{-1}(x),~ f:\mathbb R \rightarrow \mathbb{R}$,
– dxiv
Commented Aug 18, 2022 at 5:55
• Both of them increase, so I think it’s fine. @dxiv Commented Aug 18, 2022 at 6:20
• @insipidintegrator It would be fine if you spelled that out in the answer, and proved it or at least linked to a proof.
– dxiv
Commented Aug 18, 2022 at 6:41

$$f(x)=x^9+3x^6+3x^3-16x+9$$

$$f(x)=0\Rightarrow (x-1)(x^2+x-1)(x^6+2x^4+2x^3+4x^2+2x+9)=0$$

where $$x^6+2x^4+2x^3+4x^2+2x+9=x^6+x^4+x^2(x+1)^2+2x^2+(x+1)^2+8>0$$ for $$\forall x\in \mathbb{R}$$, so we get:

$$(x-1)(x^2+x-1)=0$$ There are three real roots:

$$x_1=1, x_2=\frac{-1-\sqrt{5}}2, x_3=\frac{-1+\sqrt{5}}2$$

• Is there a way you can justify how factored the 8th-degree polynomial? Commented Aug 17, 2022 at 3:44
• either use long division, or add/subtract terms to extract the $x^2+x-1$ factor. Commented Aug 17, 2022 at 3:47