Question

if x+y+z=1 and xy +yz + zx =-1 and xyz = -1 then find x³+y³+z³

Answer 1

$$\begin{gathered} \mathrm{Use}\mathrm{the}\mathrm{algebraic}\mathrm{formula}. \\ {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3-3\mathrm{xyz}=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-\mathrm{xy}-\mathrm{yz}-\mathrm{zx}\right) \\ \mathrm{This}\mathrm{implies}\mathrm{that}, \\ {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3=\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2-\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\right)+3\mathrm{xyz}....\left(1\right) \\ \mathrm{We}\mathrm{know}\mathrm{the}\mathrm{algebraic}\mathrm{formula}. \\ {\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^2={\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2+2\mathrm{xy}+2\mathrm{yz}+2\mathrm{zx} \\ \mathrm{This}\mathrm{gives}, \\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^2-2\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ \mathrm{Since}\mathrm{x}+\mathrm{y}+\mathrm{z}=1,\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=-1,\mathrm{we}\mathrm{get} \\ {\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\left(1\right)}^2-2\left(-1\right) \\ =1+2 \\ =3 \\ \mathrm{Then}\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ {\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3=\left(1\right)\left(3-\left(-1\right)\right)+3\left(-1\right) \\ =\left(1\right)\left(3+1\right)-3 \\ =\left(1\right)\left(4\right)-3 \\ =4-3 \\ =\mathbf{1} \end{gathered}$$