Solve erify: x ^ 3 + y ^ 3 + z ^ 3 - 3xyz = 1/2 * (x + y + z)\{(x - y) ^ 2 + (y - z) ^ 2 + (z - x) ^ 2\}

Question

$$x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = \frac { 1 } { 2 } ( x + y + z ) [ ( x - y ) ^ { 2 } + ( y - z ) ^ { 2 } + ( x - x ^ { 2 } ) ^ { 2 } ]$$

Solve for y (complex solution)

$\left\{\begin{matrix}\\y=-x-z\text{, }&\text{unconditionally}\\y\in \mathrm{C}\text{, }&x=-\sqrt{z}\text{ or }x=\sqrt{z}\text{ or }x=\sqrt{1-z}+1\text{ or }x=-\sqrt{1-z}+1\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-y\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+y^{2}+\left(y-z\right)^{2}+\left(x-x^{2}\right)^{2}\right)$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(y-z\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+y^{2}+y^{2}-2yz+z^{2}+\left(x-x^{2}\right)^{2}\right)$$

Combine $y^{2}$ and $y^{2}$ to get $2y^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+\left(x-x^{2}\right)^{2}\right)$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-x^{2}\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2xx^{2}+\left(x^{2}\right)^{2}\right)$$

To multiply powers of the same base, add their exponents. Add $1$ and $2$ to get $3$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2x^{3}+\left(x^{2}\right)^{2}\right)$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2x^{3}+x^{4}\right)$$

Combine $x^{2}$ and $x^{2}$ to get $2x^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}\right)$$

Use the distributive property to multiply $\frac{1}{2}$ by $x+y+z$.

$$x^{3}+y^{3}+z^{3}-3xyz=\left(\frac{1}{2}x+\frac{1}{2}y+\frac{1}{2}z\right)\left(2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}\right)$$

Use the distributive property to multiply $\frac{1}{2}x+\frac{1}{2}y+\frac{1}{2}z$ by $2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}$ and combine like terms.

$$x^{3}+y^{3}+z^{3}-3xyz=x^{3}-2xyz+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $2xyz$ to both sides.

$$x^{3}+y^{3}+z^{3}-3xyz+2xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine $-3xyz$ and $2xyz$ to get $-xyz$.

$$x^{3}+y^{3}+z^{3}-xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $y^{3}$ from both sides.

$$x^{3}+y^{3}+z^{3}-xyz-y^{3}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine $y^{3}$ and $-y^{3}$ to get $0$.

$$x^{3}+z^{3}-xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $\frac{1}{2}yz^{2}$ to both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $yx^{3}$ to both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $\frac{1}{2}yx^{4}$ from both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $x^{3}$ from both sides.

$$z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}-x^{3}$$

Combine $x^{3}$ and $-x^{3}$ to get $0$.

$$z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $z^{3}$ from both sides.

$$-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}-z^{3}$$

Combine $\frac{1}{2}z^{3}$ and $-z^{3}$ to get $-\frac{1}{2}z^{3}$.

$$-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}-\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine all terms containing $y$.

$$\left(-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}\right)y=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}-\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

The equation is in standard form.

$$\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)y=\frac{zx^{4}}{2}-zx^{3}+\frac{x^{5}}{2}+\frac{xz^{2}}{2}+zx^{2}-x^{4}-\frac{z^{3}}{2}$$

Divide both sides by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

$$\frac{\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)y}{-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}}=\frac{\left(x+z\right)\left(x^{2}-z\right)\left(x^{2}-2x+z\right)}{2\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)}$$

Dividing by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$ undoes the multiplication by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

$$y=\frac{\left(x+z\right)\left(x^{2}-z\right)\left(x^{2}-2x+z\right)}{2\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)}$$

Divide $\frac{\left(-2x+x^{2}+z\right)\left(x+z\right)\left(x^{2}-z\right)}{2}$ by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

$$y=-\left(x+z\right)$$

Show Solution

Solve for y

$\left\{\begin{matrix}\\y=-x-z\text{, }&\text{unconditionally}\\y\in \mathrm{R}\text{, }&\left(x=\sqrt{1-z}+1\text{ and }z\leq 1\right)\text{ or }\left(x=-\sqrt{1-z}+1\text{ and }z\leq 1\right)\text{ or }\left(z\geq 0\text{ and }|x|=\sqrt{z}\right)\end{matrix}\right.$

Steps for Solving Linear Equation

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-y\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+y^{2}+\left(y-z\right)^{2}+\left(x-x^{2}\right)^{2}\right)$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(y-z\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+y^{2}+y^{2}-2yz+z^{2}+\left(x-x^{2}\right)^{2}\right)$$

Combine $y^{2}$ and $y^{2}$ to get $2y^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+\left(x-x^{2}\right)^{2}\right)$$

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(x-x^{2}\right)^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2xx^{2}+\left(x^{2}\right)^{2}\right)$$

To multiply powers of the same base, add their exponents. Add $1$ and $2$ to get $3$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2x^{3}+\left(x^{2}\right)^{2}\right)$$

To raise a power to another power, multiply the exponents. Multiply $2$ and $2$ to get $4$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(x^{2}-2xy+2y^{2}-2yz+z^{2}+x^{2}-2x^{3}+x^{4}\right)$$

Combine $x^{2}$ and $x^{2}$ to get $2x^{2}$.

$$x^{3}+y^{3}+z^{3}-3xyz=\frac{1}{2}\left(x+y+z\right)\left(2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}\right)$$

Use the distributive property to multiply $\frac{1}{2}$ by $x+y+z$.

$$x^{3}+y^{3}+z^{3}-3xyz=\left(\frac{1}{2}x+\frac{1}{2}y+\frac{1}{2}z\right)\left(2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}\right)$$

Use the distributive property to multiply $\frac{1}{2}x+\frac{1}{2}y+\frac{1}{2}z$ by $2x^{2}-2xy+2y^{2}-2yz+z^{2}-2x^{3}+x^{4}$ and combine like terms.

$$x^{3}+y^{3}+z^{3}-3xyz=x^{3}-2xyz+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $2xyz$ to both sides.

$$x^{3}+y^{3}+z^{3}-3xyz+2xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine $-3xyz$ and $2xyz$ to get $-xyz$.

$$x^{3}+y^{3}+z^{3}-xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+y^{3}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $y^{3}$ from both sides.

$$x^{3}+y^{3}+z^{3}-xyz-y^{3}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine $y^{3}$ and $-y^{3}$ to get $0$.

$$x^{3}+z^{3}-xyz=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-\frac{1}{2}yz^{2}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $\frac{1}{2}yz^{2}$ to both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}-yx^{3}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Add $yx^{3}$ to both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+\frac{1}{2}yx^{4}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $\frac{1}{2}yx^{4}$ from both sides.

$$x^{3}+z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $x^{3}$ from both sides.

$$z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=x^{3}+\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}-x^{3}$$

Combine $x^{3}$ and $-x^{3}$ to get $0$.

$$z^{3}-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Subtract $z^{3}$ from both sides.

$$-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}+\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}-z^{3}$$

Combine $\frac{1}{2}z^{3}$ and $-z^{3}$ to get $-\frac{1}{2}z^{3}$.

$$-xyz+\frac{1}{2}yz^{2}+yx^{3}-\frac{1}{2}yx^{4}=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}-\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

Combine all terms containing $y$.

$$\left(-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}\right)y=\frac{1}{2}xz^{2}-x^{4}+\frac{1}{2}x^{5}+zx^{2}-\frac{1}{2}z^{3}-zx^{3}+\frac{1}{2}zx^{4}$$

The equation is in standard form.

$$\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)y=\frac{zx^{4}}{2}-zx^{3}+\frac{x^{5}}{2}+\frac{xz^{2}}{2}+zx^{2}-x^{4}-\frac{z^{3}}{2}$$

Divide both sides by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

Dividing by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$ undoes the multiplication by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

$$y=\frac{\left(x+z\right)\left(x^{2}-z\right)\left(x^{2}-2x+z\right)}{2\left(-\frac{x^{4}}{2}-xz+x^{3}+\frac{z^{2}}{2}\right)}$$

Divide $\frac{\left(-2x+x^{2}+z\right)\left(x+z\right)\left(x^{2}-z\right)}{2}$ by $-xz+\frac{1}{2}z^{2}+x^{3}-\frac{1}{2}x^{4}$.

$$y=-\left(x+z\right)$$

Show Solution

Solve for x (complex solution)

$x=-\sqrt{1-z}+1$
$x=-\left(y+z\right)$
$x=\sqrt{1-z}+1$
$x=\sqrt{z}$
$x=-\sqrt{z}$

Solve for x

$\left\{\begin{matrix}\\x=-\left(y+z\right)\text{, }&\text{unconditionally}\\x=\sqrt{1-z}+1\text{; }x=-\sqrt{1-z}+1\text{, }&z\leq 1\\x=\sqrt{z}\text{; }x=-\sqrt{z}\text{, }&z\geq 0\end{matrix}\right.$

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