Solve 14f(x)=x^-3+2x

$$14 f ( x ) = x ^ { - 3 } + 2 x$$

Solve for f

$$f=\frac{1}{7}+\frac{1}{14x^{4}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

The equation is in standard form.

$$14xf=2x+\frac{1}{x^{3}}$$

Divide both sides by 14x.

$$\frac{14xf}{14x}=\frac{2x+\frac{1}{x^{3}}}{14x}$$

Dividing by 14x undoes the multiplication by 14x.

$$f=\frac{2x+\frac{1}{x^{3}}}{14x}$$

Divide 2x+\frac{1}{x^{3}} by 14x.

$$f=\frac{1}{7}+\frac{1}{14x^{4}}$$

Solve for x (complex solution)

$$x=\frac{2^{\frac{3}{4}}i\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=\frac{2^{\frac{3}{4}}\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=-\frac{2^{\frac{3}{4}}\left(7f-1\right)^{-\frac{1}{4}}}{2}$$

$$x=-\frac{2^{\frac{3}{4}}i\left(7f-1\right)^{-\frac{1}{4}}}{2}\text{, }f\neq \frac{1}{7}$$

Solve for x

$$x=\frac{2^{\frac{3}{4}}}{2\sqrt[4]{7f-1}}$$

$$x=-\frac{2^{\frac{3}{4}}}{2\sqrt[4]{7f-1}}\text{, }f>\frac{1}{7}$$