Solve the equation $(2-x^3)^3+x-2=0$

Question

I would like for you to help me solve this equation using simple factorisation.

$(2-x^3)^3+x-2=0, x \in \mathbb{R}$

I have been trying to expand the expression which now becomes $-x^9+6x^6-12x^3+x+6$ and find a way around it but still no luck. Any help would be appreciated.

@PeterAllen: $$ = (2-x^3)^3+x-2=0 = -(x - 1) (x^2 + x + 2) (x^6 - x^4 - 4 x^3 + x^2 + 2 x + 3) = 0$$ — Moo, Feb 01 '21 at 19:29
Following @TitoEliatron's observation, we need only check$$x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6>0$$for all $x\in\Bbb R$. Amazingly, this was a recent question. — J.G., Feb 01 '21 at 19:29
The same question was asked by me actually looking to modify/create a question for some young students by really just creating an exercise out of a polynomial I saw in a school textbook. However, these students have only been exposed to simple factorisation, but I'm not sure it can be done with elementary methods without observing that $x=1$. — Peter Allen, Feb 01 '21 at 19:34

Jessie · Accepted Answer · 2021-02-01T20:49:04.843

0

$$\begin{aligned} (2−x^3)^3+x−2=0 \\ x^9+6x^6-12x^3+x+6=0 \\ -\left(x-1\right)\left(x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6\right)=0 \\ \mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0 \\x-1=0\quad \mathrm{or}\quad \:x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0 \\x-1=0 \\ \boxed{x=1} \end{aligned}$$

edited Feb 01 '21 at 20:49

answered Feb 01 '21 at 20:09

Jessie

1,463

How do you get the idea of dividing by $x-1$? Does it not presuppose that you know the root? – Peter Allen Feb 01 '21 at 20:38
1

@Peter Allan It should be obvious that $x=1$ is a solution: $(2-1)^3 +1-2 =1+1-2$. In such examples it is always good to look if there is something obvious. – Oбжорoв Feb 01 '21 at 20:50
Indeed, I agree. – Peter Allen Feb 01 '21 at 21:00

score 0 · Answer 2 · answered Aug 28 '21 at 23:59

The function $ \ g(x) = (2 - x^3)^3 \ $ is a very steeply-decreasing function, except near two locations, as indicated by its derivative $ \ g'(x) \ = \ -9x^2 · (2 - x^3)^2 \ = \ -9 · [x · (2-x^3)]^2 \ : $ one of those is the neighborhood of its $ \ y-$ intercept at $ \ (0 \ , \ 8) \ \ , $ the other is the neighborhood of its $ \ x-$ intercept, $ \ ( \sqrt[3]2 \ , \ 0 ) \ \ . $ Examination of the second derivative, $ \ g''(x) \ = \ -36x · (2x^6 - 5x^3 + 2) \ $ indicates that both of these are inflection points (upward to downward concavity). This is also suggested for $ \ x = \ \sqrt[3]2 \ $ by the fact that it is a triple zero of $ \ g(x) \ \ . $ What is important to note is that the function curve has no "turning-points".

By writing the given equation as the equation for the intersection of two function curves, $ \ (2 - x^3)^3 \ = \ 2 - x \ \ , $ it is perhaps a bit easier to see that $ \ x = 1 \ $ is a solution. We find that $ \ g'(1) \ = \ -9 \ $ and has similarly large negative slope in the vicinity of $ \ ( 1 \ , \ 1) \ \ . $ The intersection of the curve for $ \ g(x) \ $ with the line $ \ y \ = \ 2 - x \ $ is therefore unique. There is then only a single real solution for the equation $ \ (2 - x^3)^3 + x - 2 \ = \ 0 \ \ . $ (The disposition of its eight complex roots is discussed in the comments and the linked page.)

Solve the equation $(2-x^3)^3+x-2=0$

2 Answers2