Solve for x 9x^5=81x^3

$9 x^{5} = 81 x^{3}$

Step 1

Subtract $81 x^{3}$ from both sides of the equation.

$9 x^{5} - 81 x^{3} = 0$

Step 2

Factor the left side of the equation.

Factor $9 x^{3}$ out of $9 x^{5} - 81 x^{3}$ .

Factor $9 x^{3}$ out of $9 x^{5}$ .

$9 x^{3} (x^{2}) - 81 x^{3} = 0$

Factor $9 x^{3}$ out of $- 81 x^{3}$ .

$9 x^{3} (x^{2}) + 9 x^{3} (- 9) = 0$

Factor $9 x^{3}$ out of $9 x^{3} (x^{2}) + 9 x^{3} (- 9)$ .

$9 x^{3} (x^{2} - 9) = 0$

Rewrite $9$ as $3^{2}$ .

$9 x^{3} (x^{2} - 3^{2}) = 0$

Factor.

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 $b = 3$ .

$9 x^{3} ((x + 3) (x - 3)) = 0$

Remove unnecessary parentheses.

$9 x^{3} (x + 3) (x - 3) = 0$

Step 3

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

$x^{3} = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 4

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

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

$x^{3} = 0$

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

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 7

The final solution is all the values that make $9 x^{3} (x + 3) (x - 3) = 0$ true.

$x = 0, - 3, 3$

Solve for x 9x^5=81x^3

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