Question

What is formula for ${\left(x-y-z\right)}^3$

Answer 1

Expanding expression ${\left(x-y-z\right)}^3$:

${\left(x-y-z\right)}^3$ can be rewritten as follows:

$\begin{array}{rcl}{\left(x-y-z\right)}^3& =& {\left(x-y-z\right)}^2(x-y-z)\\ & =& \left(x^2+y^2-2xy+z^2-2xz+2yz\right)(x-y-z)\mathbf{[}\mathbf{\because }{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{+}\mathbf{z}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{2}\mathbf{xz}\mathbf{+}\mathbf{2}\mathbf{yz}\mathbf{]}\\ & =& \left(x^3-y^3-z^3+x{y}^2-2x^{2}y+z^{2}x-2x^{2}z+2xyz-x^{2}y+2x{y}^2-z^{2}y+2xyz-2y^{2}z-x^{2}z-y^{2}z+2xyz+2x{z}^2-2y{z}^2\right)\\ & =& x^3-y^3-z^3-3x^{2}y+3x{y}^2-3y^{2}z-3y{z}^2+3x{z}^2-3x^{2}z+6xyz\end{array}$.

Thus, the formula for ${\left(x-y-z\right)}^{\mathbf{3}}$ is.$\begin{array}{rcl}& & {\mathbf{x}}^{\mathbf{3}}\mathbf{-}{\mathbf{y}}^{\mathbf{3}}\mathbf{-}{\mathbf{z}}^{\mathbf{3}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}{\mathbf{xy}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{z}\mathbf{-}\mathbf{3}{\mathbf{yz}}^{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{xz}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{z}\mathbf{+}\mathbf{6}\mathbf{xyz}\end{array}$