Question

Solve for z in equation (z-1)^3+z^3=0​

Answer 1

`z = (1)/(2) \\ z = (1 + i√(3) )/(2) \\ z = (1 - i√(3) )/(2)`

where the last two are needed only over the complex

Explanation:

first of all, we can see thay is the sum of two cubes, whixh has a known factorization

`{x}^(3) + {y}^(3) = (x + y)( {x}^(2) - xy + {y}^(2) )`

in our case, the expression becomes

`((z - 1) + z)( {(z - 1)}^(2) - z(z - 1) + {z}^(2) )`

at this point, since if a product is zero either factor needs to be zero:

• if you are to solve over the real numbers, the only factor that contibutes is the first, which gives you

`z = (1)/(2)`

and you're done;

• if you want complex solution, we need to handle the second part. it's simple algebra, we just multiply to get
• 
  `{z}^(2) - 2z + 1 - {z}^(2) + z + {z}^(2) = 0 \\ {z }^(2) - z + 1 = 0 \\ z = (1 \pm i√(3) )/(2)`

Answer 2

z= 1

Explanation:

`(z-1)^(3)` + z^3= 0

`z^(3) - 3z^(2) +3z-1 + z^(3) =0`

2z^3 - 3z^2 + 3z - 1 = 0

z(2z^2 - 3z +3) - 1 = 0

z= 1 or 2z^2 - 3z +3= 0

z∈IR

∴z= 1