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Find the following products and verify the result for $x = -1, y = -2$:
$(x^2y-1) (3-2x^2y)$
Given:
$(x^2y-1) (3-2x^2y)$
To do:
We have to find the given product and verify the result for $x = -1, y = -2$.
Solution:
We know that,
$(a+b)\times(c+d)=a(c+d)+b(c+d)$
Therefore,
$(x^2y-1) (3-2x^2y)=x^2y(3 - 2x^2y) -1(3-2x^2y)$
$= x^2y(3) - x^2y(2x^2y)- 1(3)+ 1(2x^2y)$
$= 3x^2y-2x^{2+2}y^{1+1}-3+ 2x^2y$
$= 3x^2y-2x^4y^2-3+2x^2y$
$= 3x^2y + 2x^2y - 2x^4y^2 - 3$
$= 5x^2y - 2x^4y^2 - 3$
LHS $= (x^2y - 1) (3 - 2x^2y)$
$= [(-1)^2(-2) -1] [3 - 2 (-1)^2 (-2)]$
$= [1 (-2) -1) [3 - 2(1) (-2)]$
$= (-2 - 1) (3 + 4)$
$= -3(7)$
$= -21$
RHS $= 5x^2y - 2x^4y^2 - 3$
$= 5(-1)^2 (-2) -2 (-1)^4 (-2)^2 -3$
$=5 (1) (-2) - 2 (1) (4) -3$
$= -10-8-3$
$= -21$
Therefore,
LHS $=$ RHS
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