Question

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

Answer 1

$3x^3{y}^2\times (2x-3y)$

$=3x^3{y}^2\cdot 2x-3x^3{y}^2\cdot 3y$

$=6x^4{y}^2-9x^3{y}^3$

Now, for $x=-1,y=2$, we have

L.H.S.-

$3x^3{y}^2\times (2x-3y)$

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

$=3\cdot (-1)\cdot 4\times (-2-6)$

$=(-12)\times (-8)$

$=96$

R.H.S.-

$6x^4{y}^2-9x^3{y}^3$

$=6{(-1)}^4{\left(2\right)}^2-9{(-1)}^3{\left(2\right)}^3$

$=6\cdot 1\cdot 4-9\cdot (-1)\cdot 8$

$=24+72$

$=96$

$\because$ L.H.S. $=$ R.H.S.

Hence verified.