98. Two cubic equations. Let $a, b, c$ be the roots of the equation $x^{3} + q x + r = 0$. Write the equation whose roots are the numbers

$$
\frac{b+c}{a^{2}}, \quad \frac{c+a}{b^{2}}, \quad \frac{a+b}{c^{2}}
$$
98. Since the coefficient of $x^{2}$ in the original equation is 0, it follows that $a+b+c=0$. Therefore, $b+c=-a, c+a=-b$ and $a+b=-c$. In other words, we need to find the equation whose roots are $-\frac{1}{a},-\frac{1}{b}$ and $-\frac{1}{c}$ - the reciprocals of the roots of the original equation, taken with the opposite sign. Therefore, we should simply write the coefficients of the original equation in reverse order and change the sign of those that stand at even powers of $x$. As a result, we get the desired equation

$$
r x^{3}-q x^{2}-1=0
$$

[A. Wayne, S. S. M., 48, 492 (June 1948).]