Question

How (x³+y³)+z³-3xyz = [(x+y)³-3xy(x+y)]+z³-3xyz

Answer 1

$$\begin{gathered} Asweknowthat, \\ {\left(x+y\right)}^3=x^3+y^3+3xy\left(x+y\right) \\ {x}^3+y^3={\left(x+y\right)}^3-3xy\left(x+y\right) \\ Soputtingthisvaluein, \\ \left(x^3+y^3\right)+z^3-3xyzweget, \\ =\left[{\left(x+y\right)}^3-3xy\left(x+y\right)\right]+z^3-3xyz \end{gathered}$$