Find the Roots (Zeros) y^3-27=9y^2-27y

$y^{3} - 27 = 9 y^{2} - 27 y$

Step 1

Solve for $y$ .

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$9 y^{2} - 27 y = y^{3} - 27$

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

$9 y^{2} - 27 y - y^{3} = - 27$

Add $27$ to both sides of the equation.

$9 y^{2} - 27 y - y^{3} + 27 = 0$

Factor the left side of the equation.

Reorder terms.

$- y^{3} + 9 y^{2} - 27 y + 27 = 0$

Rewrite $- y^{3} + 9 y^{2} - 27 y + 27$ in a factored form.

Regroup terms.

$- y^{3} + 27 + 9 y^{2} - 27 y = 0$

Factor $- 1$ out of $- y^{3} + 27$ .

Factor $- 1$ out of $- y^{3}$ .

$- (y^{3}) + 27 + 9 y^{2} - 27 y = 0$

Rewrite $27$ as $- 1 (- 27)$ .

$- (y^{3}) - 1 \cdot - 27 + 9 y^{2} - 27 y = 0$

Factor $- 1$ out of $- (y^{3}) - 1 (- 27)$ .

$- (y^{3} - 27) + 9 y^{2} - 27 y = 0$

Rewrite $27$ as $3^{3}$ .

$- (y^{3} - 3^{3}) + 9 y^{2} - 27 y = 0$

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 $a = y$ and $b = 3$ .

$- ((y - 3) (y^{2} + y \cdot 3 + 3^{2})) + 9 y^{2} - 27 y = 0$

Factor.

Simplify.

Move $3$ to the left of $y$ .

$- ((y - 3) (y^{2} + 3 y + 3^{2})) + 9 y^{2} - 27 y = 0$

Raise $3$ to the power of $2$ .

$- ((y - 3) (y^{2} + 3 y + 9)) + 9 y^{2} - 27 y = 0$

Remove unnecessary parentheses.

$- (y - 3) (y^{2} + 3 y + 9) + 9 y^{2} - 27 y = 0$

Factor $9 y$ out of $9 y^{2} - 27 y$ .

Factor $9 y$ out of $9 y^{2}$ .

$- (y - 3) (y^{2} + 3 y + 9) + 9 y (y) - 27 y = 0$

Factor $9 y$ out of $- 27 y$ .

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

Factor $9 y$ out of $9 y (y) + 9 y (- 3)$ .

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

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

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

$(y - 3) (- 1 (y^{2} + 3 y + 9)) + 9 y (y - 3) = 0$

Factor $y - 3$ out of $9 y (y - 3)$ .

$(y - 3) (- 1 (y^{2} + 3 y + 9)) + (y - 3) (9 y) = 0$

Factor $y - 3$ out of $(y - 3) (- 1 (y^{2} + 3 y + 9)) + (y - 3) (9 y)$ .

$(y - 3) (- 1 (y^{2} + 3 y + 9) + 9 y) = 0$

Apply the distributive property.

$(y - 3) (- 1 y^{2} - 1 (3 y) - 1 \cdot 9 + 9 y) = 0$

Simplify.

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

$(y - 3) (- y^{2} - 1 (3 y) - 1 \cdot 9 + 9 y) = 0$

Multiply $3$ by $- 1$ .

$(y - 3) (- y^{2} - 3 y - 1 \cdot 9 + 9 y) = 0$

Multiply $- 1$ by $9$ .

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

Add $- 3 y$ and $9 y$ .

$(y - 3) (- y^{2} + 6 y - 9) = 0$

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 9 = 9$ and whose sum is $b = 6$ .

Factor $6$ out of $6 y$ .

$(y - 3) (- y^{2} + 6 (y) - 9) = 0$

Rewrite $6$ as $3$ plus $3$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

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

Factor the polynomial by factoring out the greatest common factor, $- y + 3$ .

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

Remove unnecessary parentheses.

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

Combine exponents.

Factor $- 1$ out of $y$ .

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

Rewrite $- 3$ as $- 1 (3)$ .

$(- 1 (- y) - 1 \cdot 3) (- y + 3) (y - 3) = 0$

Factor $- 1$ out of $- 1 (- y) - 1 (3)$ .

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

Raise $- y + 3$ to the power of $1$ .

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

Raise $- y + 3$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Factor $- 1$ out of $- y$ .

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

Rewrite $3$ as $- 1 (- 3)$ .

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

Factor $- 1$ out of $- (y) - 1 (- 3)$ .

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

Rewrite $- (y - 3)$ as $- 1 (y - 3)$ .

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

Apply the product rule to $- 1 (y - 3)$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply ${(y - 3)}^{2}$ by $1$ .

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

Raise $y - 3$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

Divide each term in $- 1 {(y - 3)}^{3} = 0$ by $- 1$ and simplify.

Divide each term in $- 1 {(y - 3)}^{3} = 0$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide ${(y - 3)}^{3}$ by $1$ .

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

Simplify the right side.

Divide $0$ by $- 1$ .

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

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

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

Step 2

Set $9 y^{2} - 27 y$ equal to $0$ .

$3 = 0$

Step 3

Since $3 \neq 0$ , there are no solutions.

No solution

Find the Roots (Zeros) y^3-27=9y^2-27y

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