(i) (a + b)3 – a – b
(ii) x2 – 2xy + y2 – a2 – 2ab – b2
Answer:
(i) (a + b)3 – a – b
Above terms can be written as,
(a + b)3 – (a + b)
Take out common in all terms,
(a + b) [(a + b)2 – 1]
(a + b) [(a + b)2 – 12]
We know that, a2 – b2 = (a + b) (a – b)
(a + b) (a + b + 1) (a + b – 1)
(ii) x2 – 2xy + y2 – a2 – 2ab – b2
x2 – 2xy + y2 – a2 – 2ab – b2
Above terms can be written as,
(x2 – 2xy + y2) – (a2 + 2ab + b2)
We know that, (a + b)2 = a2 + 2ab + b2 and (a – b)2 = a2 – 2ab + b2
(x2 – (2 × x × y) + y2) – (a2 + (2 × a × b) + b2)
(x – y)2 – (a + b)2
We know that, a2 – b2 = (a + b) (a – b)
[(x – y) + (a + b)] [(x – y) – (a + b)]
(x – y + a + b) (x – y – a – b)
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