Verify:
(i) 
(ii) 
(i) 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) 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)
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