Y^6+124y^3-125... i give best answers?

Question

factor the polynomial as much as possible...

i give best answers

Answer 1

(y³+125)(y³-1)

(y+5)(y²-5y+25)(y-1)(y²+y+1)

The quadratics here do not have real solutions.

Answer 2

First think of it as a a quadratic in y^3:

y^6 + 124y^3 - 125 = (y^3 + 125)(y^3 - 1)

Now factor each of the two factors on the right hand side of the above equation using the sum/difference of cubes formulae:

y^3 + 125 = (y + 5)(y^2 - 5y + 25)

y^3 - 1 = (y - 1)(y^2 + y + 1)

Both of the quadratic factors in the above two lines are irreducible over the rationals (as well as over the reals), so in all:

y^6 + 124y^3 - 125 = (y + 5)(y - 1)(y^2 - 5y + 25)(y^2 + y + 1)

Answer 3

y^6+124y^3-125

(y^3+125)(y^3-1)

(y^3+125)(y-1)(y^2+y+1)

(y+5)(y^2-5y+25)(y-1)(y^2+y+1)