If $A = 7 x^{2} + 5 x y - 9 y^{2}, B = - 4 x^{2} + x y + 5 y^{2}$ and $C = 4 y^{2} - 3 x^{2} - 6 x y$ then show that $A + B + C = 0.$

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Solution

Given:
$A$ $= 7 x^{2} + 5 x y - 9 y^{2}$
$B$ $= 4 x^{2} + x y + 5 y^{2}$
$C$ $= 4 y^{2} - 3 x^{2} - 6 x y$

Substitute the value of $A, B$ and $C$ in $A + B + C,$
$A + B + C =$ $(7 x^{2} + 5 x y - 9 y^{2}) + (- 4 x^{2} + x y + 5 y^{2}) + (4 y^{2} - 3 x^{2} - 6 x y)$
$= 7 x^{2} + 5 x y - 9 y^{2} - 4 x^{2} + x y + 5 y^{2} + 4 y^{2} - 3 x^{2} - 6 x y$
$= 7 x^{2} - 4 x^{2} - 3 x^{2} + 5 x y + x y - 6 x y - 9 y^{2} + 5 y^{2} + 4 y^{2}$
$= (7 - 4 - 3) x^{2} + (5 + 1 - 6) x y - (9 - 5 - 4) y^{2}$
$= 0 x^{2} + 0 x y - 0 y^{2}$
$= 0$

Hence, proved $A + B + C = 0 .$

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A = 7 x^{2} + 5 x y - 9 y^{2}, B = - 4 x^{2} + x y + 5 y^{2}

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