Solve k^2x(3)^3-kx(3)^2+3kx3-k=0

$$k ^ { 2 } x ( 3 ) ^ { 3 } - k x ( 3 ) ^ { 2 } + 3 k x 3 - k = 0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{3x_{3}-1}{9\left(3k-1\right)}\text{, }&k\neq \frac{1}{3}\\x\in \mathrm{C}\text{, }&k=0\text{ or }\left(x_{3}=\frac{1}{3}\text{ and }k=\frac{1}{3}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 3 to the power of 3 and get 27.

$$k^{2}x\times 27-kx\times 3^{2}+3kx_{3}-k=0$$

Calculate 3 to the power of 2 and get 9.

$$k^{2}x\times 27-kx\times 9+3kx_{3}-k=0$$

Subtract 3kx_{3} from both sides. Anything subtracted from zero gives its negation.

$$k^{2}x\times 27-kx\times 9-k=-3kx_{3}$$

Add k to both sides.

$$k^{2}x\times 27-kx\times 9=-3kx_{3}+k$$

Multiply -1 and 9 to get -9.

$$k^{2}x\times 27-9kx=-3kx_{3}+k$$

Combine all terms containing x.

$$\left(k^{2}\times 27-9k\right)x=-3kx_{3}+k$$

The equation is in standard form.

$$\left(27k^{2}-9k\right)x=k-3kx_{3}$$

Divide both sides by 27k^{2}-9k.

$$\frac{\left(27k^{2}-9k\right)x}{27k^{2}-9k}=\frac{k-3kx_{3}}{27k^{2}-9k}$$

Dividing by 27k^{2}-9k undoes the multiplication by 27k^{2}-9k.

$$x=\frac{k-3kx_{3}}{27k^{2}-9k}$$

Divide -3kx_{3}+k by 27k^{2}-9k.

$$x=\frac{1-3x_{3}}{9\left(3k-1\right)}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{3x_{3}-1}{9\left(3k-1\right)}\text{, }&k\neq \frac{1}{3}\\x\in \mathrm{R}\text{, }&k=0\text{ or }\left(x_{3}=\frac{1}{3}\text{ and }k=\frac{1}{3}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

Calculate 3 to the power of 3 and get 27.

$$k^{2}x\times 27-kx\times 3^{2}+3kx_{3}-k=0$$

Calculate 3 to the power of 2 and get 9.

$$k^{2}x\times 27-kx\times 9+3kx_{3}-k=0$$

Subtract 3kx_{3} from both sides. Anything subtracted from zero gives its negation.

$$k^{2}x\times 27-kx\times 9-k=-3kx_{3}$$

Add k to both sides.

$$k^{2}x\times 27-kx\times 9=-3kx_{3}+k$$

Multiply -1 and 9 to get -9.

$$k^{2}x\times 27-9kx=-3kx_{3}+k$$

Combine all terms containing x.

$$\left(k^{2}\times 27-9k\right)x=-3kx_{3}+k$$

The equation is in standard form.

$$\left(27k^{2}-9k\right)x=k-3kx_{3}$$

Divide both sides by 27k^{2}-9k.

$$\frac{\left(27k^{2}-9k\right)x}{27k^{2}-9k}=\frac{k-3kx_{3}}{27k^{2}-9k}$$

Dividing by 27k^{2}-9k undoes the multiplication by 27k^{2}-9k.

$$x=\frac{k-3kx_{3}}{27k^{2}-9k}$$

Divide -3kx_{3}+k by 27k^{2}-9k.

$$x=\frac{1-3x_{3}}{9\left(3k-1\right)}$$

Solve for k (complex solution)

$$\left\{\begin{matrix}\\k=0\text{, }&\text{unconditionally}\\k=-\frac{-9x+3x_{3}-1}{27x}\text{, }&x\neq 0\\k\in \mathrm{C}\text{, }&x=0\text{ and }x_{3}=\frac{1}{3}\end{matrix}\right.$$

Solve for k

$$\left\{\begin{matrix}\\k=0\text{, }&\text{unconditionally}\\k=-\frac{-9x+3x_{3}-1}{27x}\text{, }&x\neq 0\\k\in \mathrm{R}\text{, }&x=0\text{ and }x_{3}=\frac{1}{3}\end{matrix}\right.$$