Find the following product and verify the result for x = − 1, y = − 2:

To multiply, we will  use distributive law as follows:

$\left(x^{2} y-1\right)\left(3-2 x^{2} y\right)$

$=x^{2} y\left(3-2 x^{2} y\right)-1 \times\left(3-2 x^{2} y\right)$

$=3 x^{2} y-2 x^{4} y^{2}-3+2 x^{2} y$

$=5 x^{2} y-2 x^{4} y^{2}-3$

$\therefore\left(x^{2} y-1\right)\left(3-2 x^{2} y\right)=5 x^{2} y-2 x^{4} y^{2}-3$

Now, we put $x=-1$ and $y=-2$ on both sides to verify the result.

$\operatorname{LHS}=\left(x^{2} y-1\right)\left(3-2 x^{2} y\right)$

$=\left[(-1)^{2}(-2)-1\right]\left[3-2(-1)^{2}(-2)\right]$

$=[1 \times(-2)-1][3-2 \times 1 \times(-2)]$

$=(-2-1)(3+4)$

$=-3 \times 7$

$=-21$

$\mathrm{RHS}=5 x^{2} y-2 x^{4} y^{2}-3$

$=5(-1)^{2}(-2)-2(-1)^{4}(-2)^{2}-3$

$=[5 \times 1 \times(-2)]-[2 \times 1 \times 4]-3$

$=-10-8-3$

$=-21$

Because LHS is equal to RHS, the result is verified.

Thus, the answer is $5 x^{2} y-2 x^{4} y^{2}-3$.