Solve for x 9^6=x^3*9^3

$9^{6} = x^{3} \cdot 9^{3}$

Rewrite the equation as $x^{3} \cdot 9^{3} = 9^{6}$ .

$x^{3} \cdot 9^{3} = 9^{6}$

Divide each term in $(x^{3} \cdot 9^{3}) = 9^{6}$ by $9^{3}$ and simplify.

Divide each term in $(x^{3} \cdot 9^{3}) = 9^{6}$ by $9^{3}$ .

$\frac{(x^{3} \cdot 9^{3})}{9^{3}} = \frac{9^{6}}{9^{3}}$

Simplify the left side.

Cancel the common factor of $9^{3}$ .

Cancel the common factor.

$\frac{x^{3} \cdot 9^{3}}{9^{3}} = \frac{9^{6}}{9^{3}}$

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{9^{6}}{9^{3}}$

Simplify the right side.

Cancel the common factor of $9^{6}$ and $9^{3}$ .

Factor $9^{3}$ out of $9^{6}$ .

$x^{3} = \frac{9^{3} \cdot 9^{3}}{9^{3}}$

Cancel the common factors.

Multiply by $1$ .

$x^{3} = \frac{9^{3} \cdot 9^{3}}{9^{3} \cdot 1}$

Cancel the common factor.

$x^{3} = \frac{9^{3} \cdot 9^{3}}{9^{3} \cdot 1}$

Rewrite the expression.

$x^{3} = \frac{9^{3}}{1}$

Divide $9^{3}$ by $1$ .

$x^{3} = 9^{3}$

Raise $9$ to the power of $3$ .

$x^{3} = 729$

Subtract $729$ from both sides of the equation.

$x^{3} - 729 = 0$

Factor the left side of the equation.

Rewrite $729$ as $9^{3}$ .

$x^{3} - 9^{3} = 0$

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 $b = 9$ .

$(x - 9) (x^{2} + x \cdot 9 + 9^{2}) = 0$

Simplify.

Move $9$ to the left of $x$ .

$(x - 9) (x^{2} + 9 x + 9^{2}) = 0$

Raise $9$ to the power of $2$ .

$(x - 9) (x^{2} + 9 x + 81) = 0$

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

$x - 9 = 0$

$x^{2} + 9 x + 81 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Set $x^{2} + 9 x + 81$ equal to $0$ and solve for $x$ .

Set $x^{2} + 9 x + 81$ equal to $0$ .

$x^{2} + 9 x + 81 = 0$

Solve $x^{2} + 9 x + 81 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 9$ , and $c = 81$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 81)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 81}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 81}}{2 \cdot 1}$

Multiply $- 4$ by $81$ .

$x = \frac{- 9 \pm \sqrt{81 - 324}}{2 \cdot 1}$

Subtract $324$ from $81$ .

$x = \frac{- 9 \pm \sqrt{- 243}}{2 \cdot 1}$

Rewrite $- 243$ as $- 1 (243)$ .

$x = \frac{- 9 \pm \sqrt{- 1 \cdot 243}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (243)}$ as $\sqrt{- 1} \cdot \sqrt{243}$ .

$x = \frac{- 9 \pm \sqrt{- 1} \cdot \sqrt{243}}{2 \cdot 1}$

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

$x = \frac{- 9 \pm i \cdot \sqrt{243}}{2 \cdot 1}$

Rewrite $243$ as $9^{2} \cdot 3$ .

Factor $81$ out of $243$ .

$x = \frac{- 9 \pm i \cdot \sqrt{81 (3)}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$x = \frac{- 9 \pm i \cdot \sqrt{9^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 9 \pm i \cdot (9 \sqrt{3})}{2 \cdot 1}$

Move $9$ to the left of $i$ .

$x = \frac{- 9 \pm 9 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 9 \pm 9 i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{9 - 9 i \sqrt{3}}{2}, - \frac{9 + 9 i \sqrt{3}}{2}$

The final solution is all the values that make $(x - 9) (x^{2} + 9 x + 81) = 0$ true.

$x = 9, - \frac{9 - 9 i \sqrt{3}}{2}, - \frac{9 + 9 i \sqrt{3}}{2}$

Solve for x 9^6=x^3*9^3

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