Find the product of $3 x^{3} y^{2}$ and $(2 x - 3 y)$ . Also, verify the result for $x = - 1, y = 2$ .

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Solution

$3 x^{3} y^{2} \times (2 x - 3 y)$
$= 3 x^{3} y^{2} \cdot 2 x - 3 x^{3} y^{2} \cdot 3 y$
$= 6 x^{4} y^{2} - 9 x^{3} y^{3}$

Now, for $x = - 1, y = 2$ , we have
L.H.S.-
$3 x^{3} y^{2} \times (2 x - 3 y)$
$= 3 {(- 1)}^{3} {(2)}^{2} (2 \cdot (- 1) - 3 \cdot (2))$
$= 3 \cdot (- 1) \cdot 4 \times (- 2 - 6)$
$= (- 12) \times (- 8)$
$= 96$

R.H.S.-
$6 x^{4} y^{2} - 9 x^{3} y^{3}$
$= 6 {(- 1)}^{4} {(2)}^{2} - 9 {(- 1)}^{3} {(2)}^{3}$
$= 6 \cdot 1 \cdot 4 - 9 \cdot (- 1) \cdot 8$
$= 24 + 72$
$= 96$

$∵$ L.H.S. $=$ R.H.S.
Hence verified.

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