Multiply $-\frac{3}{2}x^2y^3$ by $(2x-y)$ and verify the answer for $x = 1$ and $y = 2$.
To do:
We have to multiply $-\frac{3}{2}x^2y^3$ by $(2x-y)$ and verify the answer for $x = 1$ and $y = 2$.
Solution:
$\frac{-3}{2} x^{2} y^{3} \times(2 x-y)=\frac{-3}{2} x^{2} y^{3} \times 2 x-\frac{3}{2} x^{2} y^{3} \times(-y)$
$=\frac{-3}{2} \times 2 \times x^{2+1} y^{3}-\frac{3}{2} \times(-y) x^{2} y^{3}$
$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{3+1}$
$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$
If $x=1, y=2$, then
LHS $=\frac{-3}{2} x^{2} y^{3} \times(2 x-y)$
$=\frac{-3}{2}(1)^{2}(2)^{3}(2 \times 1-2)$
$=\frac{-3}{2} \times 1 \times 8 \times 0$
$=0$
RHS $=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$
$=-3(1)^{3}(2)^{3}+\frac{3}{2}(1)^{2}+(2)^{4}$
$=-3 \times 1 \times 8+\frac{3}{2} \times 1 \times 16$
$=-24+24$
$=0$
Therefore,
LHS $=$ RHS
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