Question

Verify $x^3-y^3=(x-y)(x^2+xy+y^2)$ using some non-zero positive integers and check by actual multiplication. Can you call theses as identities?

Answer 1

To prove: $x^3-y^3=(x-y)(x^2+xy+y^2)$

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

$(x-y)(x^2+xy+y^2)=x^3+x^{2}y+x{y}^2-y{x}^2-x{y}^2-y^3=(x^3-y^3)+(x^{2}y+x{y}^2+x^{2}y-x{y}^2)=x^3-y^3=LHS$

Hence proved.

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

Let us take $x=2$ and $y=1$ in $x^3-y^3=(x-y)(x^2+xy+y^2)$ then the $LHS$ and $RHS$ will be equal as shown below:

$2^3-1^3=7$ and

$(2-1)(2^2+(2\times 1)+1^2)=1(5+2)=1\times 7=7$

Therefore, $LHS=RHS$

Hence, $x^3-y^3=(x-y)(x^2+xy+y^2)$ can be used as an identity.