how do i solve this equation? 3y/y+2 + 72/ Y^3+8 = 24/y^2-2y+4algebra

3y/(y+2) + 72/(y^3 + 8) = 24/(y^2 - 2y +
4))

1st y^3 + 8 = (y + 2)(y^2 - 2y + 4)   (sum of
squares)
so

3y/(y+2) + 72/((y+2)(y^2 - 2y + 4)) =
24/(y^2 - 2y + 4)  
Multiply everythin by (y+2)(y^2-2y+4) and we
get

3y(y^2-2y+4) + 72 = 24(y+2)
so

3y^3 - 6y^2 + 12y + 72 = 24y + 48, now put in standard
form

3y^3 - 6y^2 - 12y + 24 = 0  and we can divide by 3 to
get

y^3 - 2y^2 - 4y + 8 = 0   which has -2, 2 as
solutions

y^2(y - 2) - 4(y - 2) = 0 so we
get

(y-2)(y^2 - 4) = 0 and
finally

(y-2)(y-2)(y+2) = 0  so y = -2 or y =
2.

Now this is important, we need to check these
solutions.

3y/(y+2) + 72/(y^3+8) = 24/(y^2 - 2y +
4)

When we substitute y = -2 into this
equation
3(-2)/(-2+2) + 72/((-2)^3+8) = 24/((-2)^2 - 2(-2) + 4) we
get
-6/0 + 72/0 = 24/(12)
we get division by zero so y = -2 is an
extraneous solution, but when we substitute y = 2 we
get

3(2)/(2+2) + 72/(2^3+8) = 24/(2^2 - 2(2) +
4)

6/4 + 72/16 = 24/(4)
6/4 + 18/4 = 24/4  which
is an identity, so
The only solution is y = 2.