Question

verify x^3+y^3=(x+y)(x^2-xy+y^2)​

Answer 1

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x³−y³=(x−y)(x²+xy+y²)

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✒ x³−y³=(x−y)(x²+xy+y²)

Consider the right hand side (RHS) and expand it as follows:

(x−y)(x² +xy+y²)=x³+x²y+xy² −yx² −xy²−y³

=(x³−y³)+(x² y+xy² +x² y−xy² )=x³ −y³

=LHS

Hence proved.

Yes, we can call it as an identity: For example:

Let us take x=2 and y=1 in x³-y³

=(x−y)(x²+xy+y²) then the LHS and RHS will be equal as shown below:

2³-1³=7

(2-1)(2²+(2×1)+1² )=1(5+2)=1×7=7

7 = 7

Therefore, LHS=RHS

Hence, x³−y³=(x−y)(x²+xy+y²) can be used as an identity.

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Answer 2

R.H.S => (x+y) ( x2 - xy + y2)

=> x3 - x2y + xy2 + x2y - xy2 + y3              

{On multiplying x3 + y3 with (x+y) ( x2 - xy + y2)}

=> [x3 + y3] + ( -x2y + x2y) + ( xy2 - xy2)

=>  x3 + y3

Since R.H.S = L.H.S,

That is x3 + y3 = x3 + y3

Hence, verified that

x3 + y3 = (x+y) ( x2 - xy + y2)