Question

Verify: ${\mathrm{x}}^3-{\mathrm{y}}^3=(\mathrm{x}-\mathrm{y})({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2)$

Answer 1

We have,

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

In the given equation

$$\begin{gathered} \mathrm{LHS}={\mathrm{x}}^3-{\mathrm{y}}^3 \\ \mathrm{RHS}=(\mathrm{x}-\mathrm{y})({\mathrm{x}}^2+\mathrm{xy}+{\mathrm{y}}^2) \end{gathered}$$

Consider LHS,

We know that,

$$\begin{gathered} {(\mathrm{x}-\mathrm{y})}^3={\mathrm{x}}^3-{\mathrm{y}}^3-3\mathrm{xy}(\mathrm{x}-\mathrm{y}) \\ \Rightarrow {\mathrm{x}}^3-{\mathrm{y}}^3={(\mathrm{x}-\mathrm{y})}^3+3\mathrm{xy}(\mathrm{x}+\mathrm{y}) \end{gathered}$$

$$\begin{gathered} \mathrm{Taking}(\mathrm{x}-\mathrm{y})\mathrm{common}\mathrm{in}\mathrm{RHS} \\ {\mathrm{x}}^3-{\mathrm{y}}^3=(\mathrm{x}-\mathrm{y})[{(\mathrm{x}-\mathrm{y})}^2+3\mathrm{xy}] \\ \Rightarrow {\mathrm{x}}^3-{\mathrm{y}}^3=(\mathrm{x}-\mathrm{y})\left[\right({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy})+3\mathrm{xy}] \\ \Rightarrow {\mathrm{x}}^3-{\mathrm{y}}^3=(\mathrm{x}-\mathrm{y})({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{xy}) \\ \mathrm{LHS}=\mathrm{RHS} \end{gathered}$$

Hence Verified.