$Factorise the given expression: a b (x^{2} + y^{2}) + x y (a^{2} + b^{2})$

(b x + a y) (a x + b y)

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(a x + b y)

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(b x^{2} + a y^{2}) (a x + b y)

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(b + a) (a x + b y)

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Solution

The correct option is A $(b x + a y) (a x + b y)$
$Given expression: a b (x^{2} + y^{2}) + x y (a^{2} + b^{2})$

$By applying the distributivity property we get,$
$= a b x^{2} + a b y^{2} + a^{2} x y + b^{2} x y$

$By regrouping the terms we get,$
$= (a b x^{2} + a^{2} x y) + (a b y^{2} + b^{2} x y)$

$By taking the common factors out, we get,$
$= a x (b x + a y) + b y (a y + b x) = (b x + a y) (a x + b y)$

$Thus, a b (x^{2} + y^{2}) + x y (a^{2} + b^{2}) = (b x + a y) (a x + b y) .$

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(b x - a y) (a x - b y)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1, - \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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