$\triangle$ Transform the equation:

$x^{2}(x-3)^{2}-16(x-3)^{2}+9 x^{2}=0$, $x^{2}\left((x-3)^{2}+9\right)-16(x-3)^{2}=0$, $x^{2}\left(x^{2}-6 x+18\right)-16(x-3)^{2}=0$, $x^{4}-6 x^{2}(x-3)-16(x-3)^{2}=0$.

Divide the last equation by $(x-3)^{2}$ (check that no solutions are lost):

$$
\left(\frac{x^{2}}{x-3}\right)^{2}-6 \frac{x^{2}}{x-3}-16=0
$$

Let $\frac{x^{2}}{x-3}=y$. Then

$$
y^{2}-6 y-16=0 ; \quad y_{1}=8, \quad y_{2}=-2
$$

Now it is not difficult to find $x$.

Answer: $-1 \pm \sqrt{7}$.
276. Solve the equation:

$$
\left(x^{2}-16\right)(x-3)^{2}+9 x^{2}=0
$$