the question is (z-a)^3=8
it is known that z1*z2*z3=-9
One way to approach the problem is to solve the first equation to find all possible values for (z-a). Since (z-a) is raised to the third power in this equation, we expect to find three possible values. This will give you three equations all of the form z-a=#.
I'm not completely sure if I understand what you mean by the second equation, but I'm assuming that in your notation z1, z2, and z3 are the three possible values for z. So the three equations you got earlier are z1-a=#, z2-a=#, and z3-a=# (# stands for three different numbers). I'm also assuming that a is some complex constant. (Some particular number of the form x+y*i.)
Using the three equations you got from solving the first equation and also using z1*z2*z3=-9 you can solve for z1, z2, z3, and a. This is because you now have four equations and four unknowns:
z1-a=#
z2-a=#
z3-a=#
z1*z2*z3=-9
the original (z-a)^3=8 has 3 roots
z1 z2 z3
i found them but transforming the equation to a trigonometric form
and then i made DEMUAVER law and found these 3 roots
then i multiplied them and i could not solve the equetion wich came out
I presume that when you used d'Moivre's formula you found that
z- a= 2, or [itex]-1+ i\sqrt{3}[/itex], or [itex]-1-i\sqrt{3}[/itex].
Then z= 2+ a, or [itex]a-1+i\sqrt{3}[/itex], or [itex]a-1- i\sqrt{3}[/itex].
Multiplying those together and setting them equal to -9, you get a very simple equation for a!
The equation you have on your file is exactly what I got: (a+2)(a2- 2a+ 4)= -9. Did you multiply it out? It is exactly a3+ 8= -9 or a3= -17. Surely you can solve that- just take the cube root.
