Q15 of 65 Page 137

Let us factorise the following algebraic expressions:

(x2 – y2) (a2 – b2) + 4abxy

Given, (x2 – y2)(a2 – b2) + 4abxy


⇒ ((x – y)(x + y)(a – b)(a + b)) + 4abxy


⇒ (ax)2 – (bx)2 – (ay)2 + (by)2 + 4abxy


⇒ (ax)2 + (by)2 + 4abxy – (bx)2 – (ay)2


⇒ (ax)2 + (by)2 + 2abxy + 2abxy – (bx)2 – (ay)2


[from Identity I, (x + y)2 = x2 + 2xy + y2 and from identity II, (x – y)2 = x2 – 2xy + y2]


⇒ (ax + by)2 – (ay – bx)2


⇒ ((ax + by) + (ay – bx))((ax + by) – (ay – bx))


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