Algebra Examples

$\frac{x^{3} (x + 5)}{x - 9} \cdot \frac{(x - 9) (x + 8)}{3 x^{3}}$

Step 1

Set the denominator in $\frac{x^{3} (x + 5)}{x - 9}$ equal to $0$ to find where the expression is undefined.

$x - 9 = 0$

Step 2

Add $9$ to both sides of the equation.

$x = 9$

Step 3

Set the denominator in $\frac{(x - 9) (x + 8)}{3 x^{3}}$ equal to $0$ to find where the expression is undefined.

$3 x^{3} = 0$

Step 4

Solve for $x$ .

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Step 4.1

Divide each term in $3 x^{3} = 0$ by $3$ and simplify.

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Step 4.1.1

Divide each term in $3 x^{3} = 0$ by $3$ .

$\frac{3 x^{3}}{3} = \frac{0}{3}$

Step 4.1.2

Simplify the left side.

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Step 4.1.2.1

Cancel the common factor of $3$ .

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Step 4.1.2.1.1

Cancel the common factor.

$\frac{3 x^{3}}{3} = \frac{0}{3}$

Step 4.1.2.1.2

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

$x^{3} = \frac{0}{3}$

Step 4.1.3

Simplify the right side.

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Step 4.1.3.1

Divide $0$ by $3$ .

$x^{3} = 0$

Step 4.2

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

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

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

$x = 0$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = 0, x = 9$

Step 6