Solve the following equation:

$$
\sqrt[3]{2 x^{2}+x-9}+\sqrt[3]{100-x-2 x^{2}}=7
$$
Let's denote the first and second addends on the left side of (1) by $a$ and $b$. For these,

$$
a+b=7
$$

and

$$
a^{3}+b^{3}=\left(2 x^{2}+x-9\right)+\left(100-x-2 x^{2}\right)=91
$$

Using the identity $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$, we can calculate the value of $a b$ from (2) and (3):

$$
a b=\frac{(a+b)^{3}-a^{3}-b^{3}}{3(a+b)}=12
$$

Based on this and (2), the possible values of $a$ and $b$ are $a=3$ and $b=4$ or $a=4$ and $b=3$. These value pairs naturally satisfy (2) and (3) as well, so, for example, the $x$ values derived from

$$
2 x^{2}+x-9=a^{3}
$$

in (1). Choosing $a=3$ from (4),

$$
x_{1}=4 ; \quad x_{2}=-\frac{9}{2}
$$

and in the case of $a=4$,

$$
x_{3}=\frac{-1+3 \sqrt{65}}{4} ; \quad x_{4}=\frac{-1-3 \sqrt{65}}{4}
$$

Therefore, the solutions to equation (1) are precisely these.