How to solve x ^ 3 = 64

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x ^ 3 = 64 | Equatix

Question

$$x^{3}=64$$

Solve for x (complex solution)

$x=-2\sqrt{3}i-2\approx -2-3.464101615i$
$x=4$
$x=-2+2\sqrt{3}i\approx -2+3.464101615i$

Solution Steps

Subtract $64$ from both sides.

$$x^{3}-64=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $-64$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-64$ by $x-4$ to get $x^{2}+4x+16$. Solve the equation where the result equals to $0$.

$$x^{2}+4x+16=0$$

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 $a$, $4$ for $b$, and $16$ for $c$ in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{2}$$

Solve the equation $x^{2}+4x+16=0$ when $±$ is plus and when $±$ is minus.

$$x=-2i\sqrt{3}-2$$ $$x=-2+2i\sqrt{3}$$

List all found solutions.

$$x=4$$ $$x=-2i\sqrt{3}-2$$ $$x=-2+2i\sqrt{3}$$

Show Solution

Solve for x

$x=4$

Solution Steps

Subtract $64$ from both sides.

$$x^{3}-64=0$$

$$±64,±32,±16,±8,±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}-64$ by $x-4$ to get $x^{2}+4x+16$. Solve the equation where the result equals to $0$.

$$x^{2}+4x+16=0$$

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 $a$, $4$ for $b$, and $16$ for $c$ in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-48}}{2}$$

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

$$x\in \emptyset$$

List all found solutions.

$$x=4$$

Show Solution

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