Solve -f=sqrt{x^3-1}

$$- f = \sqrt { x ^ { 3 } - 1 }$$

Solve for f (complex solution)

$$f=-\sqrt{x^{3}-1}$$

Steps for Solving Linear Equation

The equation is in standard form.

$$-f=\sqrt{x^{3}-1}$$

Divide both sides by -1.

$$\frac{-f}{-1}=\frac{\sqrt{x^{3}-1}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$f=\frac{\sqrt{x^{3}-1}}{-1}$$

Divide \sqrt{x^{3}-1} by -1.

$$f=-\sqrt{x^{3}-1}$$

Solve for f

$$f=-\sqrt{x^{3}-1}$$

$$x\geq 1$$

Steps for Solving Linear Equation

The equation is in standard form.

$$-f=\sqrt{x^{3}-1}$$

Divide both sides by -1.

$$\frac{-f}{-1}=\frac{\sqrt{\left(x-1\right)\left(x^{2}+x+1\right)}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$f=\frac{\sqrt{\left(x-1\right)\left(x^{2}+x+1\right)}}{-1}$$

Divide \sqrt{\left(x-1\right)\left(x^{2}+x+1\right)} by -1.

$$f=-\sqrt{\left(x-1\right)\left(x^{2}+x+1\right)}$$

Solve for x (complex solution)

$$x=e^{\frac{2\pi i}{3}}\sqrt[3]{f^{2}+1}$$

$$x=\sqrt[3]{f^{2}+1}$$

$$x=e^{\frac{4\pi i}{3}}\sqrt[3]{f^{2}+1}\text{, }arg(f)\geq \pi \text{ or }f=0$$

Solve for x

$$x=\sqrt[3]{f^{2}+1}$$

$$f\leq 0$$