Pre-Algebra Examples

$y^{3} = 9 y$

Step 1

Subtract $9 y$ from both sides of the equation.

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

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $y$ out of $y^{3} - 9 y$ .

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Step 2.1.1

Factor $y$ out of $y^{3}$ .

$y \cdot y^{2} - 9 y = 0$

Step 2.1.2

Factor $y$ out of $- 9 y$ .

$y \cdot y^{2} + y \cdot - 9 = 0$

Step 2.1.3

Factor $y$ out of $y \cdot y^{2} + y \cdot - 9$ .

$y (y^{2} - 9) = 0$

Step 2.2

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

$y (y^{2} - 3^{2}) = 0$

Step 2.3

Factor.

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Step 2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 3$ .

$y ((y + 3) (y - 3)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$y (y + 3) (y - 3) = 0$

Step 3

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

$y = 0$

$y + 3 = 0$

$y - 3 = 0$

Step 4

Set $y$ equal to $0$ .

$y = 0$

Step 5

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

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Step 5.1

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

$y + 3 = 0$

Step 5.2

Subtract $3$ from both sides of the equation.

$y = - 3$

Step 6

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

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Step 6.1

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

$y - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$y = 3$

Step 7

The final solution is all the values that make $y (y + 3) (y - 3) = 0$ true.

$y = 0, - 3, 3$