Question 194674
First, perform synthetic division on the polynomial {{{z^3+i}}} using the test zero {{{z=i}}}. Note: you can easily solve for "z" and find that one solution is {{{z=i}}}


<pre>
i |  1  0  0   i
  |     i  -1  -i
  -----------
     1  i  -1   0
</pre>


Since the first three numbers in the third row are 1, i, and -1, this means that 


{{{z^3+i=(z-i)(z^2+iz-1)}}}



Now simply use the quadratic formula to solve {{{z^2+iz-1=0}}} for "z":



{{{z = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{z = (-(i) +- sqrt( (i)^2-4(1)(-1) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=i}}}, and {{{c=-1}}}



{{{z = (-i +- sqrt( -1-4(1)(-1) ))/(2(1))}}} Square {{{i}}} to get {{{i^2=(sqrt(-1))^2=-1}}}. 



{{{z = (-i +- sqrt( -1+4 ))/(2)}}} Multiply



{{{z = (-i +- sqrt( 3 ))/(2)}}} Combine like terms.



So the other two solutions are {{{z = (-i + sqrt( 3 ))/(2)}}} or {{{z = (-i - sqrt( 3 ))/(2)}}}



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Answer:


So the 3 solutions are {{{z=i}}}, {{{z = (-i + sqrt( 3 ))/(2)}}}, or {{{z = (-i - sqrt( 3 ))/(2)}}}