Precalculus Examples

$x^{6} - 3 x^{3} - 4 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Rewrite $x^{6}$ as ${(x^{3})}^{2}$ .

${(x^{3})}^{2} - 3 x^{3} - 4 = 0$

Step 1.2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$u^{2} - 3 u - 4 = 0$

Step 1.3

Factor $u^{2} - 3 u - 4$ using the AC method.

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Step 1.3.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Step 1.3.2

Write the factored form using these integers.

$(u - 4) (u + 1) = 0$

Step 1.4

Replace all occurrences of $u$ with $x^{3}$ .

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

Step 2

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

$x^{3} - 4 = 0$

$x^{3} + 1 = 0$

Step 3

Set $x^{3} - 4$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $x^{3} - 4$ equal to $0$ .

$x^{3} - 4 = 0$

Step 3.2

Solve $x^{3} - 4 = 0$ for $x$ .

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Step 3.2.1

Add $4$ to both sides of the equation.

$x^{3} = 4$

Step 3.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{4}$

Step 4

Set $x^{3} + 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x^{3} + 1$ equal to $0$ .

$x^{3} + 1 = 0$

Step 4.2

Solve $x^{3} + 1 = 0$ for $x$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$x^{3} = - 1$

Step 4.2.2

Add $1$ to both sides of the equation.

$x^{3} + 1 = 0$

Step 4.2.3

Factor the left side of the equation.

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Step 4.2.3.1

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

$x^{3} + 1^{3} = 0$

Step 4.2.3.2

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

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

Step 4.2.3.3

Simplify.

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Step 4.2.3.3.1

Multiply $- 1$ by $1$ .

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

Step 4.2.3.3.2

One to any power is one.

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

Step 4.2.4

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 4.2.5

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

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Step 4.2.5.1

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

$x + 1 = 0$

Step 4.2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.2.6

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

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Step 4.2.6.1

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

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

Step 4.2.6.2

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

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Step 4.2.6.2.1

Use the quadratic formula to find the solutions.

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

Step 4.2.6.2.2

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 4.2.6.2.3

Simplify.

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Step 4.2.6.2.3.1

Simplify the numerator.

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Step 4.2.6.2.3.1.1

Raise $- 1$ to the power of $2$ .

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

Step 4.2.6.2.3.1.2

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

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Step 4.2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 4.2.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 4.2.6.2.3.1.3

Subtract $4$ from $1$ .

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

Step 4.2.6.2.3.1.4

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

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

Step 4.2.6.2.3.1.5

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

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

Step 4.2.6.2.3.1.6

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

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

Step 4.2.6.2.3.2

Multiply $2$ by $1$ .

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

Step 4.2.6.2.4

The final answer is the combination of both solutions.

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

Step 4.2.7

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

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

Step 5

The final solution is all the values that make $(x^{3} - 4) (x^{3} + 1) = 0$ true.

$x = \sqrt[3]{4}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 6