Solve for y 12y^3=3y

$12 y^{3} = 3 y$

Move $3 y$ to the left side of the equation by subtracting it from both sides.

$12 y^{3} - 3 y = 0$

Factor the left side of the equation.

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Factor $3 y$ out of $12 y^{3} - 3 y$ .

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Factor $3 y$ out of $12 y^{3}$ .

$3 y (4 y^{2}) - 3 y = 0$

Factor $3 y$ out of $- 3 y$ .

$3 y (4 y^{2}) + 3 y (- 1) = 0$

Factor $3 y$ out of $3 y (4 y^{2}) + 3 y (- 1)$ .

$3 y (4 y^{2} - 1) = 0$

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

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

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

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

Factor.

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

$3 y ((2 y + 1) (2 y - 1)) = 0$

Remove unnecessary parentheses.

$3 y (2 y + 1) (2 y - 1) = 0$

Divide each term by $3$ and simplify.

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Divide each term in $3 y (2 y + 1) (2 y - 1) = 0$ by $3$ .

$\frac{3 y (2 y + 1) (2 y - 1)}{3} = \frac{0}{3}$

Simplify $\frac{3 y (2 y + 1) (2 y - 1)}{3}$ .

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Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 y (2 y + 1) (2 y - 1)}{3} = \frac{0}{3}$

Divide $(y (2 y + 1)) (2 y - 1)$ by $1$ .

$(y (2 y + 1)) (2 y - 1) = \frac{0}{3}$

Apply the distributive property.

$(y (2 y) + y \cdot 1) (2 y - 1) = \frac{0}{3}$

Simplify the expression.

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Rewrite using the commutative property of multiplication.

$(2 y \cdot y + y \cdot 1) (2 y - 1) = \frac{0}{3}$

Multiply $y$ by $1$ .

$(2 y \cdot y + y) (2 y - 1) = \frac{0}{3}$

Multiply $y$ by $y$ by adding the exponents.

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Move $y$ .

$(2 (y \cdot y) + y) (2 y - 1) = \frac{0}{3}$

Multiply $y$ by $y$ .

$(2 y^{2} + y) (2 y - 1) = \frac{0}{3}$

Expand $(2 y^{2} + y) (2 y - 1)$ using the FOIL Method.

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Apply the distributive property.

$2 y^{2} (2 y - 1) + y (2 y - 1) = \frac{0}{3}$

Apply the distributive property.

$2 y^{2} (2 y) + 2 y^{2} \cdot - 1 + y (2 y - 1) = \frac{0}{3}$

Apply the distributive property.

$2 y^{2} (2 y) + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

$2 \cdot 2 (y^{2} y) + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Multiply $y^{2}$ by $y$ by adding the exponents.

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Multiply $y^{2}$ by $y$ .

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Raise $y$ to the power of $1$ .

$2 \cdot 2 (y^{2} y^{1}) + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

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

$2 \cdot 2 y^{2 + 1} + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Add $2$ and $1$ .

$2 \cdot 2 y^{3} + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Multiply $2$ by $2$ .

$4 y^{3} + 2 y^{2} \cdot - 1 + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Multiply $- 1$ by $2$ .

$4 y^{3} - 2 y^{2} + y (2 y) + y \cdot - 1 = \frac{0}{3}$

Rewrite using the commutative property of multiplication.

$4 y^{3} - 2 y^{2} + 2 y \cdot y + y \cdot - 1 = \frac{0}{3}$

Multiply $y$ by $y$ by adding the exponents.

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Move $y$ .

$4 y^{3} - 2 y^{2} + 2 (y \cdot y) + y \cdot - 1 = \frac{0}{3}$

Multiply $y$ by $y$ .

$4 y^{3} - 2 y^{2} + 2 y^{2} + y \cdot - 1 = \frac{0}{3}$

Move $- 1$ to the left of $y$ .

$4 y^{3} - 2 y^{2} + 2 y^{2} - 1 \cdot y = \frac{0}{3}$

Rewrite $- 1 y$ as $- y$ .

$4 y^{3} - 2 y^{2} + 2 y^{2} - y = \frac{0}{3}$

Add $- 2 y^{2}$ and $2 y^{2}$ .

$4 y^{3} + 0 - y = \frac{0}{3}$

Add $4 y^{3}$ and $0$ .

$4 y^{3} - y = \frac{0}{3}$

Divide $0$ by $3$ .

$4 y^{3} - y = 0$

Factor the left side of the equation.

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Factor $y$ out of $4 y^{3} - y$ .

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Factor $y$ out of $4 y^{3}$ .

$y (4 y^{2}) - y = 0$

Factor $y$ out of $- y$ .

$y (4 y^{2}) + y \cdot - 1 = 0$

Factor $y$ out of $y (4 y^{2}) + y \cdot - 1$ .

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

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$y ({(2 y)}^{2} - 1) = 0$

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

$y ({(2 y)}^{2} - 1^{2}) = 0$

Factor.

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

$y ((2 y + 1) (2 y - 1)) = 0$

Remove unnecessary parentheses.

$y (2 y + 1) (2 y - 1) = 0$

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

$y = 0$

$2 y + 1 = 0$

$2 y - 1 = 0$

Set the first factor equal to $0$ .

$y = 0$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 y + 1 = 0$

Subtract $1$ from both sides of the equation.

$2 y = - 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 y = - 1$ by $2$ .

$\frac{2 y}{2} = \frac{- 1}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 y}{2} = \frac{- 1}{2}$

Divide $y$ by $1$ .

$y = \frac{- 1}{2}$

Move the negative in front of the fraction.

$y = - \frac{1}{2}$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 y - 1 = 0$

Add $1$ to both sides of the equation.

$2 y = 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 y = 1$ by $2$ .

$\frac{2 y}{2} = \frac{1}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 y}{2} = \frac{1}{2}$

Divide $y$ by $1$ .

$y = \frac{1}{2}$

The final solution is all the values that make $y (2 y + 1) (2 y - 1) = 0$ true.

$y = 0, - \frac{1}{2}, \frac{1}{2}$

The result can be shown in multiple forms.

Exact Form:

$y = 0, - \frac{1}{2}, \frac{1}{2}$

Decimal Form:

$y = 0, - 0.5, 0.5$

Solve for y 12y^3=3y

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