Question

Verify \[{{\text{x}}^3} - {{\text{y}}^3} = \left( {{\text{x - y}}} \right)\left( {{{\text{x}}^2} + {\text{xy + }}{{\text{y}}^2}} \right)\]using some non-zero positive integers and check by actual multiplication. Can you call these as identities?

Answer 1

Hint: In order to prove, we take L.H.S use multiplication and prove that is equal to R.H.S. Take random numbers for x and y to prove it as an identity.

Step-by-step answer:
To prove: \[{{\text{x}}^3} - {{\text{y}}^3} = \left( {{\text{x - y}}} \right)\left( {{{\text{x}}^2} + {\text{xy + }}{{\text{y}}^2}} \right)\]

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

\[\left( {{\text{x - y}}} \right)\left( {{{\text{x}}^2} + {\text{xy + }}{{\text{y}}^2}} \right)\]= ${{\text{x}}^3} + {{\text{x}}^2}{\text{y + x}}{{\text{y}}^2} - {\text{y}}{{\text{x}}^2} - {\text{x}}{{\text{y}}^2} - {{\text{y}}^3}$
                                         = $\left( {{{\text{x}}^3} - {{\text{y}}^3}} \right) + \left( {{{\text{x}}^2}{\text{y + x}}{{\text{y}}^2} - {\text{y}}{{\text{x}}^2} - {\text{x}}{{\text{y}}^2}} \right)$
                                         = \[{{\text{x}}^3} - {{\text{y}}^3}\]= LHS

Hence, proved.

Yes, we can call it an identity.

For example:
Let us take x = 2 and y = 1 in \[{{\text{x}}^3} - {{\text{y}}^3} = \left( {{\text{x - y}}} \right)\left( {{{\text{x}}^2} + {\text{xy + }}{{\text{y}}^2}} \right)\]
Then the LHS and RHS will be equal as shown below:

L.H.S = 8-1
           = 7
R.H.S = $\left( {2 - 1} \right)$ $\left( {{2^2} + \left( {2 \times 1} \right) + {1^2}} \right)$
           = 1(5 + 2)
           = 1 × 7 = 7

Therefore, LHS = RHS

Hence, \[{{\text{x}}^3} - {{\text{y}}^3} = \left( {{\text{x - y}}} \right)\left( {{{\text{x}}^2} + {\text{xy + }}{{\text{y}}^2}} \right)\] can be used as an identity.

Note: In order to solve this type of question the key is to take either L.H.S or R.H.S and convert it into the other. We can prove this as an identity by assuming different random numbers for x and y and solving the equation, it comes out to be true for all values of x and y.