Solve by Factoring x^6-1=0

$x^{6} - 1 = 0$

Step 1

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

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

Step 2

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

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

Step 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 $a = x^{2}$ and $b = 1$ .

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

Step 4

Simplify.

Rewrite $1$ as $1^{2}$ .

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

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

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

Multiply $x^{2}$ by $1$ .

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

Step 5

Simplify each term.

Multiply the exponents in ${(x^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Multiply $2$ by $2$ .

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

One to any power is one.

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

Step 6

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

$x + 1 = 0$

$x - 1 = 0$

$x^{4} + x^{2} + 1 = 0$

Step 7

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 8

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 9

Set $x^{4} + x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{4} + x^{2} + 1$ equal to $0$ .

$x^{4} + x^{2} + 1 = 0$

Solve $x^{4} + x^{2} + 1 = 0$ for $x$ .

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

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

$u = x^{2}$

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 $u$ .

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

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

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

$u = \frac{- 1 \cdot 1 + i \sqrt{3}}{2}$

Factor $- 1$ out of $i \sqrt{3}$ .

$u = \frac{- 1 \cdot 1 - (- i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

$u = \frac{- 1 (1 - i \sqrt{3})}{2}$

Move the negative in front of the fraction.

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $1$ .

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

Subtract $4$ from $1$ .

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

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

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

Factor $- 1$ out of $- i \sqrt{3}$ .

$u = \frac{- 1 \cdot 1 - (i \sqrt{3})}{2}$

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

$u = \frac{- 1 (1 + i \sqrt{3})}{2}$

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

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

Substitute the real value of $u = x^{2}$ back into the solved equation.

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

${(x^{2})}^{1} = - \frac{1 + i \sqrt{3}}{2}$

Solve the first equation for $x$ .

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

Solve the equation for $x$ .

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

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

Simplify $\pm \sqrt{- \frac{1 - i \sqrt{3}}{2}}$ .

Rewrite $- \frac{1 - i \sqrt{3}}{2}$ as $i^{2} \frac{1^{2} - i \sqrt{3}}{2}$ .

Rewrite $- 1$ as $i^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Rewrite $\sqrt{\frac{1 - i \sqrt{3}}{2}}$ as $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ .

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

Multiply $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

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

Combine $i$ and $\frac{\sqrt{(1 - i \sqrt{3}) \cdot 2}}{2}$ .

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

Move $2$ to the left of $1 - i \sqrt{3}$ .

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Next, use the negative value of the $\pm$ to find the second solution.

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

The complete solution is the result of both the positive and negative portions of the solution.

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

Solve the second equation for $x$ .

${(x^{2})}^{1} = - \frac{1 + i \sqrt{3}}{2}$

Solve the equation for $x$ .

Remove parentheses.

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

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

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

Simplify $\pm \sqrt{- \frac{1 + i \sqrt{3}}{2}}$ .

Rewrite $- \frac{1 + i \sqrt{3}}{2}$ as $i^{2} \frac{1^{2} + i \sqrt{3}}{2}$ .

Rewrite $- 1$ as $i^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Rewrite $\sqrt{\frac{1 + i \sqrt{3}}{2}}$ as $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ .

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

Multiply $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$x = \pm i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Combine using the product rule for radicals.

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

Combine $i$ and $\frac{\sqrt{(1 + i \sqrt{3}) \cdot 2}}{2}$ .

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

Move $2$ to the left of $1 + i \sqrt{3}$ .

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Next, use the negative value of the $\pm$ to find the second solution.

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

The complete solution is the result of both the positive and negative portions of the solution.

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

The solution to $x^{4} + x^{2} + 1 = 0$ is $x = \frac{i \sqrt{2 (1 - i \sqrt{3})}}{2}, - \frac{i \sqrt{2 (1 - i \sqrt{3})}}{2}, \frac{i \sqrt{2 (1 + i \sqrt{3})}}{2}, - \frac{i \sqrt{2 (1 + i \sqrt{3})}}{2}$ .

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

Step 10

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

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

Solve by Factoring x^6-1=0

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