Solve the equation $x^{6}-6 x+5=0$.
I. solution. Let's plot the function $f(x)=x^{6}-6 x+5$ using a computer program.

Based on the graph, it can be guessed that the function has a root, possibly a multiple root, at $x=1$. Let's check the value of the function at $x=1$: $f(1)=0$. Indeed, $x=1$ is a root of the equation, so we can factor out $(x-1)$ from the polynomial:

$$
\begin{aligned}
f(x) & =x^{6}-6 x+5= \\
& =(x-1)\left(x^{5}+x^{4}+x^{3}+x^{2}+x-5\right)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_ac50874cf2737a8fed53g-1.jpg?height=591&width=439&top_left_y=520&top_left_x=838)

If we substitute $x=1$ into the second factor, we again get 0. Therefore, we can factor out $x-1$ once more:

$$
f(x)=(x-1)^{2}\left(x^{4}+2 x^{3}+3 x^{2}+4 x+5\right)
$$

Here, the last factor does not yield 0 at $x=1$. Based on the graph, we suspect that there are no more roots. If we can show that the last factor is always positive, we will have proven that the equation has no more roots.

$x^{4}+2 x^{3}$ resembles the square of $x^{2}+x$: $\left(x^{2}+x\right)^{2}=x^{4}+2 x^{3}+x^{2}$. There remains $2 x^{2}+4 x+5$, which can be easily transformed:

$$
2 x^{2}+4 x+5=2\left(x^{2}+2 x+1\right)+3=2(x+1)^{2}+3
$$

We have obtained that

$$
x^{4}+2 x^{3}+3 x^{2}+4 x+5=\left(x^{2}+x\right)^{2}+2(x+1)^{2}+3 \geq 3
$$

This proves that the equation has only one root at $x=1$.

II. solution. Notice that $x=1$ is a solution to the equation. We will show that this is the only root. Let's examine the derivative of the function $f(x)=x^{6}-6 x+5$. $f^{\prime}(x)=6 x^{5}-6$. This is only 0 when $x=1$. The derivative is positive for $x>1$. Therefore, $f(x)$ is strictly decreasing when $x<1$ and strictly increasing when $x>1$. This implies that $f(x)>0$ when $x \neq 1$.

Since the function is only 0 at $x=1$, the polynomial has this as its only root.

Remark. Several people incorrectly interpreted the zero derivative as equivalent to the existence of an extremum, whereas it is only true that if an extremum exists, then the derivative is 0.

III. solution. Rearranging the equation, we get $x^{6}+5=6 x$.

The left side of this equation is at least 5 when $x \leq 0$, while the right side is at most 0. Therefore, there are no roots in this case.

Divide both sides of the obtained equation by 6. Using the inequality between the arithmetic and geometric means for $x>0$, we get:

$$
\frac{x^{6}+5}{6}=\frac{x^{6}+1+1+1+1+1}{6} \geq \sqrt[6]{x^{6} \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1}=x
$$

where equality holds when $x^{6}=1$, i.e., for $x>0$, when $x=1$.

Thus, the only real solution to the equation is $x=1$.