Solve by Factoring x^6+6x^3+5=0

$x^{6} + 6 x^{3} + 5 = 0$

Step 1

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

${(x^{3})}^{2} + 6 x^{3} + 5 = 0$

Step 2

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

$u^{2} + 6 u + 5 = 0$

Step 3

Factor $u^{2} + 6 u + 5$ using the AC method.

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 $5$ and whose sum is $6$ .

$1, 5$

Write the factored form using these integers.

$(u + 1) (u + 5) = 0$

Step 4

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

$(x^{3} + 1) (x^{3} + 5) = 0$

Step 5

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

$(x^{3} + 1^{3}) (x^{3} + 5) = 0$

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 $b = 1$ .

$(x + 1) (x^{2} - x \cdot 1 + 1^{2}) (x^{3} + 5) = 0$

Step 7

Simplify.

Multiply $- 1$ by $1$ .

$(x + 1) (x^{2} - x + 1^{2}) (x^{3} + 5) = 0$

One to any power is one.

$(x + 1) (x^{2} - x + 1) (x^{3} + 5) = 0$

Step 8

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

$x + 1 = 0$

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

$x^{3} + 5 = 0$

Step 9

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 10

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

Step 11

Set $x^{3} + 5$ equal to $0$ and solve for $x$ .

Set $x^{3} + 5$ equal to $0$ .

$x^{3} + 5 = 0$

Solve $x^{3} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{3} = - 5$

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

$x = \sqrt[3]{- 5}$

Simplify $\sqrt[3]{- 5}$ .

Rewrite $- 5$ as ${(- 1)}^{3} \cdot 5$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x = \sqrt[3]{- 1 (5)}$

Rewrite $- 1$ as ${(- 1)}^{3}$ .

$x = \sqrt[3]{{(- 1)}^{3} \cdot 5}$

Pull terms out from under the radical.

$x = - 1 \sqrt[3]{5}$

Rewrite $- 1 \sqrt[3]{5}$ as $- \sqrt[3]{5}$ .

$x = - \sqrt[3]{5}$

Step 12

The final solution is all the values that make $(x + 1) (x^{2} - x + 1) (x^{3} + 5) = 0$ true.

$x = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}, - \sqrt[3]{5}$

Solve by Factoring x^6+6x^3+5=0

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