Question

Verify that $ {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right) $ .

Answer 1

Hint: We will assume x, y as any two positive numbers. Then we will substitute that numbers n to left hand side (LHS) i.e. $ {{x}^{3}}+{{y}^{3}} $ and similarly, in right hand side (RHS) i.e. $ \left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right) $ . On getting both answers we will see whether both are the same or not. Thus, we will verify from this.

Complete step-by-step answer:
Here, we have to verify the given equation $ {{x}^{3}}+{{y}^{3}}=\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right) $ . So, we will take x and y as two numbers, let’s say x as 5 and y as 2.
So, putting this value in Left hand side (LHS) i.e. $ {{x}^{3}}+{{y}^{3}} $ , we get as
 $ LHS={{\left( 5 \right)}^{3}}+{{\left( 2 \right)}^{3}} $
On solving, we get as
 $ LHS=125+8=133 $ ………………………..(1)
Similarly, we will put values in right hand side (RHS) i.e. $ \left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right) $ we get as
 $ RHS=\left( 5+2 \right)\left( {{5}^{2}}-\left( 5\cdot 2 \right)+{{2}^{2}} \right) $
On solving, we get as
 $ RHS=\left( 7 \right)\left( 25-10+4 \right) $
On further simplification, we get as
 $ RHS=\left( 7 \right)\left( 19 \right)=133 $ …………………………..(2)
Thus, we can see that equation (1) is the same as equation (2). So, $ LHS=RHS $ .
Thus, verified.

Note: Instead of taking positive numbers, we can check by taking negative integers also. If we take $ -5,-2 $ and on putting values in LHS, we get as
 $ LHS={{\left( -5 \right)}^{3}}+{{\left( -2 \right)}^{3}}=-125-8=-133 $
On putting values in RHS, we get as
 $ RHS=\left( -5-2 \right)\left( {{\left( -5 \right)}^{2}}-\left( -5\times -2 \right)+{{\left( -2 \right)}^{2}} \right) $
Thus, on solving we will get as
 $ RHS=\left( -7 \right)\left( 25+10+4 \right)=-7\times 19=-133 $
Thus, we get the same answer in this case also. Thus verified.