Calculus Examples

$4 x^{3} - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$4 x^{3} = 4$

Step 2

Subtract $4$ from both sides of the equation.

$4 x^{3} - 4 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $4$ out of $4 x^{3} - 4$ .

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Step 3.1.1

Factor $4$ out of $4 x^{3}$ .

$4 (x^{3}) - 4 = 0$

Step 3.1.2

Factor $4$ out of $- 4$ .

$4 (x^{3}) + 4 (- 1) = 0$

Step 3.1.3

Factor $4$ out of $4 (x^{3}) + 4 (- 1)$ .

$4 (x^{3} - 1) = 0$

Step 3.2

Rewrite $1$ as $1^{3}$ .

$4 (x^{3} - 1^{3}) = 0$

Step 3.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

$4 ((x - 1) (x^{2} + x \cdot 1 + 1^{2})) = 0$

Step 3.4

Factor.

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Step 3.4.1

Simplify.

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Step 3.4.1.1

Multiply $x$ by $1$ .

$4 ((x - 1) (x^{2} + x + 1^{2})) = 0$

Step 3.4.1.2

One to any power is one.

$4 ((x - 1) (x^{2} + x + 1)) = 0$

Step 3.4.2

Remove unnecessary parentheses.

$4 (x - 1) (x^{2} + x + 1) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x^{2} + x + 1 = 0$

Step 5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 6

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Step 6.2

Solve $x^{2} + x + 1 = 0$ for $x$ .

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Step 6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 6.2.2

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 6.2.3

Simplify.

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Step 6.2.3.1

Simplify the numerator.

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Step 6.2.3.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 6.2.3.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 6.2.3.1.3

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 6.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 6.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 6.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 6.2.3.2

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

Step 6.2.4

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 7

The final solution is all the values that make $4 (x - 1) (x^{2} + x + 1) = 0$ true.

$x = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$