A Cubic Equation | Problem 465

[@WindyNight114]

In the first method, looking at the system of equations, I was like oh a = 1 and b = 1 work as a solution, but I was not expecting those monstrous wolfram alpha ones.

[@bobbyheffley4955]

1/27 should be subtracted from 1/9

[@Don-Ensley]

problem
z³ + z² = -2+ 4 i
Method cubic formula.
Define c temporarily as
c = -1 + 2 i
z³ + z² -2c = 0
Let
z = u -1/3
27 u³- 9 u + 2-54c = 0
Let
u = w/3
w³ - 3 w = 54c -2
w = a+b for solution
a³ +b³ = 54c -2
ab = 1
a³ b³ = 1
b³ = 1/(a³)
a³ + 1/(a³) = 54c -2
(a³)²-(54c-2) (a³) + 1 = 0
replace c= -1+2i
a³ = { -56 + 108 i ± √ [(-56+108 i)²-4] }/2
= -28+54i ± 3√3 i√(79+112i)
w =∛ (-28-√((459√65-2133)/2)+
(54+√((459√65+2133)/2) i) +
∛ (-28+√((459√65-2133)/2)+
(54-√((459√65+2133)/2) i)
= ∛[(-28+54 i)-3√(-237-336 i)] +
∛[(-28+54 i)+3√(-237-336 i)]
= 4 + 3 i
This is one of 3 roots to
w³ - 3 w + 56 - 108 i = 0
Call it r.
r = 4 + 3i.
The other 2 roots are the solutions to
w² + r w + r²-3 = 0
They are
r₂,₃ = -(4+3i)/2 ± (3i/2) √(1+8i)
Roots in w are
4 + 3 i,
-( 4 + 3i) / 2 - (3 i/2) √( 1+8i ),
-( 4 + 3i) / 2 + (3 i/2) √( 1+8i ),
back substitution:
Roots in u = w/3
4/3 + i,
-( 2/3 + i/2) - ( i/2) √( 1+8i ),
-( 2/3 + i/2) + (i/2) √( 1+8i ),
Roots in z = u - 1/3
1 + i,
-1 - i/2 - ( i/2) √( 1+8i ),
-1 - i/2 + (i/2) √( 1+8i )
answer
z ∈ { 1 + i,
-1 - i/2 - ( i/2) √( 1+8i ),
-1 - i/2 + (i/2) √( 1+8i ) }

[@NadiehFan]

Error at 4:31