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The sum of the complex roots of the equations $(x-1)^{3}+64$=0 is 

Answer:

Explanation:

We have,

  $(x-1)^{3}+64$=0 

 $\Rightarrow$           $(x-1)^{3}=-64$           

 $\Rightarrow$                $(x-1)^{3}=(-4)^{3}$

 $\Rightarrow$        x-1=-4,-4w,-4w²

 $\Rightarrow$                 x=-3,-4w+1,-4w²+1

 Complex roots of the equations are

  -4w+1,-4w²+1

Sum of complex roots are

   -4w+1-4w²+1=-4(w+w²)+2

  $=-4(-1)+2=4+2=6$    $[\because1+w+w^{2}=0]$