Question -
Answer -
(i) x3+y3 =(x+y)(x2–xy+y2)
We know that, (x+y)3 = x3+y3+3xy(x+y)
⇒ x3+y3 =(x+y)3–3xy(x+y)
⇒ x3+y3 =(x+y)[(x+y)2–3xy]
Taking (x+y) common ⇒ x3+y3 = (x+y)[(x2+y2+2xy)–3xy]
⇒ x3+y3 =(x+y)(x2+y2–xy)
(ii) x3–y3 =(x–y)(x2+xy+y2)
We know that,(x–y)3 = x3–y3–3xy(x–y)
⇒ x3−y3 =(x–y)3+3xy(x–y)
⇒ x3−y3 =(x–y)[(x–y)2+3xy]
Taking (x+y) common ⇒ x3−y3 = (x–y)[(x2+y2–2xy)+3xy]
⇒ x3+y3 =(x–y)(x2+y2+xy)