Verify: (i) x^3+y^3 = (x+y)(x^2–xy+y^2) (ii) x^3–y^3 = (x–y)(x^2+xy+y^2)

 Solutions:

(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)