**(i) ab(x**^{2} + y^{2}) – xy(a^{2} + b^{2})

^{2}+ y

^{2}) – xy(a

^{2}+ b

^{2})

**(ii) (ax + by)**^{2} + (bx – ay)^{2}

^{2}+ (bx – ay)

^{2}

**Answer :**

**(i) ab(x**^{2} + y^{2}) – xy(a^{2} + b^{2})

^{2}+ y

^{2}) – xy(a

^{2}+ b

^{2})

Above question can be written as,

abx^{2} + aby^{2} – xya^{2} – xyb^{2}

Re-arranging the above we get,

abx^{2} – xyb^{2} + aby^{2} – xya^{2}

Take out common in all terms,

bx(ax – by) + ay(by – ax)

bx(ax – by) – ay (ax – by)

(ax – by) (bx – ay)

**(ii) (ax + by)**^{2} + (bx – ay)^{2}

^{2}+ (bx – ay)

^{2}

By expanding the give question, we get,

(ax)^{2} + (by)^{2} + 2axby + (bx)^{2} + (ay)^{2} – 2bxay

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

Re-arranging the above we get,

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

Take out common in all terms,

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

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

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