# y^3-64/3y^2-17y+20

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### 1 Answer

`(y^3-64)/(3y^2 -17y +20)`

First we will factor the numerator and denominator.

==> We know that `(a^3-b^3)= (a-b)(a^2+ab+b^2) `

`==gt (y^3-64)= (y-4)(y^2 +4y +16)`

`` ==> Now we will factor the denominator.

`==gt 3y^2 -17y +20= (3y-5)(y-4).`

`` Now we will substitute.

==> `(y^3-64)/(3y^2-17y+20)= ((y-4)(y^2+4y+16))/((3y-5)(y-4))`

`` Now we will reduce (`y-4` ).

==> `(y^3-64)/(3y^2-17y+20)= (y^2+4y+16)/(3y-5).`