Solve the following equation:

$$
\frac{(x+a+b)^{7}+(x-a-b)^{7}-(x-a+b)^{7}-(x+a-b)^{7}}{(x+a+b)^{5}+(x-a-b)^{5}-(x-a+b)^{5}-(x+a-b)^{5}}=7\left(a^{2}+b^{2}\right)
$$
By introducing the abbreviations $a+b=c$ and $a-b=d$, our equation transforms as follows:

$$
\frac{\left\{(x+c)^{7}+(x-c)^{7}\right\}-\left\{(x+d)^{7}+(x-d)^{7}\right\}}{\left\{(x+c)^{5}+(x-c)^{5}\right\}-\left\{(x+d)^{5}+(x-d)^{5}\right\}}=\frac{7}{2}\left(c^{2}+d^{2}\right)
$$

Expanding the powers in the first \{\}-bracketed terms in the numerator and the denominator, the terms containing $c$ with odd exponents cancel out. The resulting expressions are:

$$
\begin{gathered}
2 x^{7}+42 x^{5} c^{2}+70 x^{3} c^{4}+14 x c^{6} \\
2 x^{5}+20 x^{3} c^{2}+10 x c^{4}
\end{gathered}
$$

The two subtrahends can be obtained by substituting $d$ for $c$ in these, so the equation becomes:

$$
\frac{42 x^{5}\left(c^{2}-d^{2}\right)+70 x^{3}\left(c^{4}-d^{4}\right)+14 x\left(c^{6}-d^{6}\right)}{20 x^{3}\left(c^{2}-d^{2}\right)+10 x\left(c^{4}-d^{4}\right)}=\frac{7}{2}\left(c^{2}+d^{2}\right) .
$$

From this, it is clear that $x=0$ does not belong to the domain of the left-hand side, and parameter value pairs for which $c^{2}=d^{2},|c|=|d|$, i.e., those in which $a=0$, and those in which $b=0$, must be excluded. The roots are thus to be calculated from the equation obtained by the appropriate simplifications:

$$
\frac{3 x^{4}+5 x^{2}\left(c^{2}+d^{2}\right)+\left(c^{4}+c^{2} d^{2}+d^{2}\right)}{10 x^{2}+5\left(c^{2}+d^{2}\right)}=\frac{1}{2}\left(c^{2}+d^{2}\right)
$$

The denominator (based on the above exclusions) is positive, so with the usual rearrangement steps and returning to the original parameters,

$$
x= \pm \sqrt[4]{\frac{3 c^{4}+8 c^{2} d^{2}+3 d^{4}}{6}}= \pm \sqrt[4]{\frac{7 a^{4}+10 a^{2} b^{2}+7 b^{4}}{3}}
$$

(two real roots) provided $a \neq b$ and $a \neq -b$. The two roots obtained would only be equal if $a=b=0$, which we have excluded.

Maria Fazekas (Pápa, Türr I. Gymnasium, 3rd grade)

Remark. The significant simplification of the left-hand side of the equation can also be anticipated from the observation that substituting $-x$ for $x$, the numerator and the denominator both change to their $-1$ multiples, and the same holds if $-a$ is substituted for $a$ and $-b$ for $b$. Therefore, in any of the cases $x=0$, $a=0$, $b=0$, the numerator and the denominator are both 0, or in other words, from the polynomial in $x$, $a$, and $b$, the common factor $a b x$ can be factored out.