Question -
Answer -
(i) x3+y3┬а=(x+y)(x2тАУxy+y2)
We know that, (x+y)3┬а= x3+y3+3xy(x+y)
тЗТ x3+y3┬а=(x+y)3тАУ3xy(x+y)
тЗТ x3+y3┬а=(x+y)[(x+y)2тАУ3xy]
Taking (x+y) common тЗТ x3+y3┬а= (x+y)[(x2+y2+2xy)тАУ3xy]
тЗТ x3+y3┬а=(x+y)(x2+y2тАУxy)
(ii) x3тАУy3┬а=(xтАУy)(x2+xy+y2)┬а
We know that,(xтАУy)3┬а= x3тАУy3тАУ3xy(xтАУy)
тЗТ x3тИТy3┬а=(xтАУy)3+3xy(xтАУy)
тЗТ x3тИТy3┬а=(xтАУy)[(xтАУy)2+3xy]
Taking (x+y) common тЗТ x3тИТy3┬а= (xтАУy)[(x2+y2тАУ2xy)+3xy]
тЗТ x3+y3┬а=(xтАУy)(x2+y2+xy)