How do you factor 27y^6-827y68?

1 Answer

Rewrite this as

27*y^6-8=(3y^2)^3-2^327y68=(3y2)323

Using the cubes difference formula

a^3-b^3=(a-b)*(a^2+ab+b^2)a3b3=(ab)(a2+ab+b2)

where a=3y^2a=3y2 and b=2b=2

we get

27y^6-8=(3 y^2 - 2) (9 y^4 + 6 y^2 + 4)27y68=(3y22)(9y4+6y2+4)

But this can be factored further as follows

27y^6-8=1/3 (sqrt(6) - 3 y) (3 y + sqrt(6)) (-3 y^2 + sqrt(6) y - 2) (3 y^2 + sqrt(6) y + 2)27y68=13(63y)(3y+6)(3y2+6y2)(3y2+6y+2)