Verify: x3+y3=(x+y)(x2–xy+y2)
We have, x3+y3=(x+y)(x2–xy+y2)
In the given equation
LHS=x3+y3RHS=(x+y)(x2–xy+y2)
We know that,
(x+y)3=x3+y3+3xy(x+y)⇒x3+y3=(x+y)3–3xy(x+y)
Taking(x+y)commoninRHSx3+y3=(x+y)[(x+y)2–3xy]⇒x3+y3=(x+y)[(x2+y2+2xy)–3xy]⇒x3+y3=(x+y)(x2+y2–xy)LHS=RHS
Hence Verified.