# Prove that the equation has only one real root.

Prove that $$(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$$ has only one real root. It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root.

But it doesn't guarantee that the real root is unique. If we calculate the root then it can be shown that it has only one real root.

• The function is strictly increasing. If $x_1 > x_0$ then $x_1 - c > x_0 -c$ and $(x_1 -c)^3 > (x_0 -c)^3$ and $\sum (x_1 -c)^3 > \sum (x_0 - c)^3$. – fleablood Nov 17 '19 at 16:05
• The real root also has multiplicity 1. – UtilityMaximiser Nov 18 '19 at 1:19

The function is strictly increasing so the function is one to one.

A solution that only uses school algebra:

Substitute $$t=x-5/2$$. Then the equation becomes $$(t+3/2)^3+(t+1/2)^3+(t-1/2)^3+(t-3/2)^3=0.$$ Expanding the brackets, we get $$4t^3+15t=0,$$ or $$t(4t^2+15)=0$$, which clearly only has one real root.

• Good and simple answer so have my +1, but do you have any intuition why one should use the transformation t=x-5/2 and not something else? – Hans Olo Nov 18 '19 at 9:13
• @Rebel-Scum To exploit the symmetry of the four terms around $x-5/2$. 5/2 is the average of 1, 2, 3, 4. – Evangelos Bampas Nov 18 '19 at 10:22
• @EvangelosBampas Indeed, but that is not mentioned in the answer and the transformation comes a bit out of the blue – Hans Olo Nov 18 '19 at 11:22

Hint:

It is the sum of four increasing functions, and $$\lim_{x\to-\infty}=-\infty$$, $$\lim_{x\to+\infty}=+\infty$$. The intermediate value theorem guarantees there is a root, and monotonicity ensures there can be no more than one.

Here is a rather primitive solution (although judging by the contents of your question you may already know this):

By symmetry, we can see that $$x=2.5$$ is a real root. We want to show that it is the only real root. Expanding the binomial cubes, we get $$4x^3-30x^2+90x-100$$ which factors into $$2(2x-5)(x^2-5x+10).$$

As the discriminant of the quadratic factor is negative, this tells us that $$x=2.5$$ is the only real root.

Hint:

Can you show the derivative is positive for all $$x$$?

Thus, the function is strictly increasing for all $$x$$.

$$x=2.5$$ is a root.

• Is the derivative really necessary here? – Bernard Nov 17 '19 at 16:10
• I suppose it is one way to show that a function is strictly increasing. – Andrew Chin Nov 17 '19 at 16:11
• and a fairly straightforward way in this case, though it's not really necessary – J. W. Tanner Nov 17 '19 at 16:28

Let $$y=\dfrac{x-1+x-2+x-3+x-4}4$$

$$x=y+2.5$$

$$(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=(y+1.5)^3+(y+.5)^3+(y-.5)^3+(y-1.5)^3$$

Now use $$(a-b)^3+(a+b)^3=2(a^3+3ab^2)$$

$$0=4y^3+6y((1.5)^2+(.5)^2)$$

The function $$x\mapsto x^3$$ is increasing, since $$x^3-y^3=(x-y)(x^2+xy+y^2)$$ and $$x^2+xy+y^2=\frac{(x+y)^2}{2}+\frac{x^2+y^2}{2}\geq 0$$.

Shifting the domain of an increasing function does not change the fact that it is increasing, so also $$x\mapsto (x-n)^3$$ is increasing for all $$n$$.

Adding increasing functions results in another increasing function.

Thus, for all natural numbers $$n$$ the polynomial $$\sum_{i=1}^n (x-i)^3$$ is increasing. Since it tends to $$\infty$$ as $$x\to\infty$$ and $$-\infty$$ as $$x\to-\infty$$, it follows that the polynomial has exactly one value at which it crosses the $$x$$-axis, that is, exactly one real root.

Here is an elementary way that uses only

• $$(1)$$: $$a^3+b^3 = (a+b)(a^2-ab+b^2)$$ and
• $$(2)$$: $$a^2+b^2 >ab$$ for $$|a|+|b| > 0$$

Note, that $$|x-1|+|x-4|\geq |x-1 +(4-x)| =3 > 0$$ and $$|x-2|+|x-3|\geq |x-2 +(3-x)| =1 > 0$$

Let's call $$p(x) = (x-1)^3+ (x-2)^3 + (x-3)^3 + (x-4)^3$$.

Now, using $$(1)$$ write $$(x-1)^3 + (x-4)^3 = (x-1 + x-4)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$ $$= (2x-5)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$ $$(x-2)^3 + (x-3)^3 = (x-2 + x-3)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$ $$= (2x-5)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$

Hence,

$$p(x) =$$ $$(2x-5)\color{blue}{\left(\underbrace{(x-1)^2 + (x-4)^2 - (x-1)(x-4)}_{\stackrel{(2)}{>}0} + \underbrace{(x-2)^2 + (x-3)^2 - (x-2)(x-3)}_{\stackrel{(2)}{>}0}\right)}$$

So,

$$p(x) = (2x-5)\color{blue}{q(x)} \mbox{ with } \color{blue}{q(x)} > 0 \mbox{ for all } x \in \mathbb{R}$$