Q. 93.8( 247 Votes )

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Answer :

(i) x3 + y3 = (x + y) (x2 + y2 – xy)

We know that,

(x + y)3 = x3 + y3 + 3xy (x + y)


⇒  x3 + y3 = (x + y)3 – 3xy (x + y)


[Taking (x + y) common]


⇒  x3 + y3 = (x + y) [(x + y)2 – 3xy)     


⇒  x3 + y3 = (x + y) [(x2 + y2 + 2xy) – 3xy]               ........[(x + y)2 = x2 + y2 + 2 x y]


⇒  x3 + y3 = (x + y) (x2 + y2 – xy)


(ii) x3 - y3 = (x - y) (x2 + y2 + xy)

We know that,


(x - y)3 = x3 - y3 - 3xy (x - y)


⇒  x3 - y3 = (x - y)3 + 3xy (x - y)



[Taking (x - y) common]

⇒  x3 - y3 = (x - y) [(x - y)2 + 3xy)


⇒ x3 - y3 = (x - y) [(x2 + y2 - 2xy) + 3xy]            ......[(x - y)2 = x2 + y2 - 2 x y]


= x3 - y3 = (x - y) (x2 + y2 + xy)

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