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QUIK ANSWER PLEASE!!! Solve the equationy^3 - 27 = 9y^2 - 27y

User RobeN
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1 Answer

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The first step is to simplify both sides of the equation. The equation can be written as

y^3 - 3^3 = 9y(y - 3)

For the left hand side, we would apply the difference of two cubes formula. it is expressed as

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

By comparing with the left hand side of the equation,

x = y and y = 3. It becomes

(y - 3)(y^2 + 3y + 3^2)

= (y - 3)(y^2 + 3y + 9)

The equation becomes

(y - 3)(y^2 + 3y + 9) = 9y(y - 3)

If we divide both sides of the equation by (y - 3), it becomes

(y - 3)(y^2 + 3y + 9)/(y - 3 = 9y(y - 3)/(y - 3)

y^2 + 3y + 9 = 9y

y^2 + 3y - 9y + 9 = 0

y^2 - 6y + 9 = 0

We would solve the quadratic equation by applying the method of factorisation. We would find two terms such that their sum or difference is - 6y and their product is 9y^2. The terms are - 3y and - 3y. The equation becomes

y^2 - 3y - 3y + 9

y(y - 3) - 3( y - 3) = 0

(y - 3)(y - 3) = 0

y - 3 = 0 twice

y = 3 twice

User Marcelus Trojahn
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