I tried to make up an equation :
$x-x^3 = 1\implies x(1-x)(1+x)=1$
Get $x=1, x=-1$. For each of the two values: $x=1$ (double roots), & $x=-1$, could not find satisfying this property.
Am confused, and am not sure if this is the way to disprove the title.
On
Denote $f(x)=x-x^3-1(x \in \mathbb{R}).$ We want to find the real roots for the equation $f(x)=0$.
Notice that $f'(x)=1-3x^2.$ Thus, we can obtain the local extremum points are $$x_1=\frac{\sqrt{3}}{3},x_2=-\frac{\sqrt{3}}{3}.$$ Based on these, we can draw the graphic of $y=f(x)$ as follows
Thus, you can see there exists only one solution for $f(x)=0$, which is located between $-2$ and $-1$.
About the principle of graphing, I give an example.
You have seen that $f'\left(-\dfrac{\sqrt{3}}{3}\right)=0$, and $f'(x)<0$ for $x<-\dfrac{\sqrt{3}}{3}$. Thus, you may know that $f(x)$ decreases with an increasing $x$,where $x \leq -\dfrac{\sqrt{3}}{3}.$ Now,you can draw a descending curve over $\left(-\infty,-\dfrac{\sqrt{3}}{3}\right).$
How about $-\dfrac{\sqrt{3}}{3}<x<\dfrac{\sqrt{3}}{3}$? And how about $x>\dfrac{\sqrt{3}}{3}$? Can you discuss the sign of $f'(x)$ and graph them?
.If $x(1-x)(1+x) = 1$, then it does not imply that $x = 1 , x=-1$ etc. In fact, it implies that $x$ cannot be equal to $1,-1$ or $0$. This is because in each of these cases the left hand side is zero, when it is supposed to be equal to one, which is the right hand side.
When the product of some numbers is zero, then at least one of them is zero. This allows us, for example when $(x-2)(x-3) = 0$, to conclude that $x=2$ or $x=3$.
However, when the product of some numbers is $1$, then nothing can be concluded about the precise value of any of the quantities. At most, you can conclude that all the terms are non-zero, so that prohibits some values of $x$, like how I prohibited above, but that still leaves you a large set of values with which to work.
Given the equation $x - x^3 = 1$, this is a cubic equation, so it can be solved by Cardano's general method for cubics. However, it turns out that $x^3-x+1$ is an irreducible polynomial (in integers) so you cannot obtain any factorization here.
What you can do, is find the location of this real root using the intermediate value theorem, (a case of) which says that if $f(x) = x^3 - x+1$ and you find two numbers $ a < b$ such that $f(a)$ and $f(b)$ have opposite signs, then there is a real root of $f$ between $a$ and $b$. Using this, we can assert the existence of a root between $-1$ and $-2$, for example, since $f(-1) = 1$ while $f(-2) = -5$, so they have differing signs.