Precalculus Examples

$x^{6} - 6 x^{3} - 7 = 0$

Step 1

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

${(x^{3})}^{2} - 6 x^{3} - 7 = 0$

Step 2

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

$u^{2} - 6 u - 7 = 0$

Step 3

Factor $u^{2} - 6 u - 7$ using the AC method.

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Step 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 $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Step 3.2

Write the factored form using these integers.

$(u - 7) (u + 1) = 0$

Step 4

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

$(x^{3} - 7) (x^{3} + 1) = 0$

Step 5

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

$(x^{3} - 7) (x^{3} + 1^{3}) = 0$

Step 6

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 $b = 1$ .

$(x^{3} - 7) ((x + 1) (x^{2} - x \cdot 1 + 1^{2})) = 0$

Step 7

Factor.

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Step 7.1

Simplify.

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Step 7.1.1

Multiply $- 1$ by $1$ .

$(x^{3} - 7) ((x + 1) (x^{2} - x + 1^{2})) = 0$

Step 7.1.2

One to any power is one.

$(x^{3} - 7) ((x + 1) (x^{2} - x + 1)) = 0$

Step 7.2

Remove unnecessary parentheses.

$(x^{3} - 7) (x + 1) (x^{2} - x + 1) = 0$

Step 8

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

$x^{3} - 7 = 0$

$x + 1 = 0$

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

Step 9

Set $x^{3} - 7$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $x^{3} - 7$ equal to $0$ .

$x^{3} - 7 = 0$

Step 9.2

Solve $x^{3} - 7 = 0$ for $x$ .

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Step 9.2.1

Add $7$ to both sides of the equation.

$x^{3} = 7$

Step 9.2.2

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

$x = \sqrt[3]{7}$

Step 10

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

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Step 10.1

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

$x + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

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

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Step 11.1

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

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

Step 11.2

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

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Step 11.2.1

Use the quadratic formula to find the solutions.

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

Step 11.2.2

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 11.2.3

Simplify.

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Step 11.2.3.1

Simplify the numerator.

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Step 11.2.3.1.1

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

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

Step 11.2.3.1.2

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

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Step 11.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 11.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 11.2.3.1.3

Subtract $4$ from $1$ .

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

Step 11.2.3.1.4

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

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

Step 11.2.3.1.5

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

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

Step 11.2.3.1.6

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

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

Step 11.2.3.2

Multiply $2$ by $1$ .

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

Step 11.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 11.2.4.1

Simplify the numerator.

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Step 11.2.4.1.1

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

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

Step 11.2.4.1.2

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

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Step 11.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 11.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 11.2.4.1.3

Subtract $4$ from $1$ .

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

Step 11.2.4.1.4

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

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

Step 11.2.4.1.5

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

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

Step 11.2.4.1.6

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

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

Step 11.2.4.2

Multiply $2$ by $1$ .

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

Step 11.2.4.3

Change the $\pm$ to $+$ .

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

Step 11.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 11.2.5.1

Simplify the numerator.

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Step 11.2.5.1.1

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

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

Step 11.2.5.1.2

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

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Step 11.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 11.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 11.2.5.1.3

Subtract $4$ from $1$ .

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

Step 11.2.5.1.4

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

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

Step 11.2.5.1.5

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

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

Step 11.2.5.1.6

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

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

Step 11.2.5.2

Multiply $2$ by $1$ .

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

Step 11.2.5.3

Change the $\pm$ to $-$ .

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

Step 11.2.6

The final answer is the combination of both solutions.

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

Step 12

The final solution is all the values that make $(x^{3} - 7) (x + 1) (x^{2} - x + 1) = 0$ true.

$x = \sqrt[3]{7}, - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$