Solve 8=v^3

$$8 = v ^ { 3 }$$

Solve for v

$$v=2$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

$$v^{3}=8$$

Subtract 8 from both sides.

$$v^{3}-8=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -8 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±8,±4,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$v=2$$

By Factor theorem, v-k is a factor of the polynomial for each root k. Divide v^{3}-8 by v-2 to get v^{2}+2v+4. Solve the equation where the result equals to 0.

$$v^{2}+2v+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 2 for b, and 4 for c in the quadratic formula.

$$v=\frac{-2±\sqrt{2^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$v=\frac{-2±\sqrt{-12}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$v\in \emptyset$$

List all found solutions.

$$v=2$$