6. For what values of $a, b$ and $c$ does the set of real roots of the equation

$$
x^{5}+2 x^{4}+a x^{2}+b x=c
$$

consist exactly of the numbers -1 and $1?$
Answer: $a=-6, b=-1, c=-4$.

Solution. Since $x=1, x=-1$ are roots of the equation $x^{5}+2 x^{4}+a x^{2}+b x-c=0$, the numbers $a, b, c$ satisfy the system

$$
\left\{\begin{array} { l } 
{ 3 + a + b - c = 0 , } \\
{ 1 + a - b - c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
b=-1, \\
c=a+2
\end{array} \Rightarrow x^{5}+2 x^{4}+a x^{2}+b x-c=\left(x^{2}-1\right)\left(x^{3}+2 x^{2}+x+2+a\right)\right.\right.
$$

The polynomial $x^{3}+2 x^{2}+x+2+a$ has at least one real root (equal to $\pm 1$):

1) if $x=1$ is a root, then $a=-6$ and $x^{3}+2 x^{2}+x+2+a=x^{3}+2 x^{2}+x-4=(x-1)\left(x^{2}+3 x+4\right)$, and since the quadratic trinomial $x^{2}+3 x+4$ has no real roots, the value $a=-6$ satisfies the conditions of the problem
2) if $x=-1$ is a root, then $a=-2$ and $x^{3}+2 x^{2}+x+2+a=x^{3}+2 x^{2}+x$, but the polynomial $x^{3}+2 x^{2}+x$ also has a root $x=0$, so the requirement of the problem is not met.