How do I solve (z-a)^3 = 8 when z1*z2*z3 = -9?

[transgalactic]

i added a file in which i tried to solve this question
the final equation does"nt come out

the question is (z-a)^3=8
it is known that z1*z2*z3=-9

i have dried to make an equation

how do i solve this equation??

please help

 

[Stainsor]

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

 

[transgalactic]

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 which came out

how do i solve this thing??

 

[HallsofIvy]

I presume that when you used d'Moivre's formula you found that
z- a= 2, or $-1+ i\sqrt{3}$, or $-1-i\sqrt{3}$.
Then z= 2+ a, or $a-1+i\sqrt{3}$, or $a-1- i\sqrt{3}$.
Multiplying those together and setting them equal to -9, you get a very simple equation for a!

 

[transgalactic]

in the end of the file i showed the equation that came out
its impossible to solve

 

[Mystic998]

It's not impossible to solve. The complex numbers are algebraically closed.

 

[HallsofIvy]

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.

 

[transgalactic]

dammmmmm
i didnt see that cube formula right under my nose
it is so simple

thank you very much