Question

how do you find the real or imaginary solutions to s^3+27=0?

Answer 1

I think in this case it would be sum of cubes \[(a+b)(a ^{2}-ab+b ^{2})\]

Answer 2

im sorry , I dont know what that means ..):

Answer 3

When you have a sum of cubes, like a^3+b^3, that factors out to \[(a+b)(a ^{2}-ab+b ^{2})\]In this problem, because s^3 and 27 are both cubes of a number, they can be factored and x can be found.

Answer 4

the answer to this problem is
\[(s+3)(s ^{2}-3s+9)=0\]You can then solve for one of the s's immediately and use the quadratic formula to find the imaginary roots