SOLUTION: If a, b, c are the zeroes of the polynomials p(x) = x³ + 4, then find the value of k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c

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Question 1145006: If a, b, c are the zeroes of the polynomials p(x) = x³ + 4, then find the value of
k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Begin with the general solution of the depressed cubic: , namely the primary (real) root



And then the two complex roots which are comprised of the primary root multiplied by the two complex cube roots of unity, namely:



In your case we have the further simplification that , so your three roots become:







The rest is just arithmetic. Ugly, nasty, unpleasant arithmetic, but arithmetic nonetheless. If I were going to actually do this arithmetic, I would let be the primary root and and be the other two. Good luck.


John

My calculator said it, I believe it, that settles it
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

            Surely and certainly,  this problem is advanced and is intended for advanced students.

            So,  I will assume that your level corresponds to the problem's level. 


(1)   = .


     It follows from the standard identity   = .


     Further, since "a" is the root of the equation   = 0,  we have  


          = 0;  hence,    =  =  = -3.   


     Therefore,   =  = .    (1)



(2)  Similarly,   = .             (2)


(3)  Therefore,


         k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c  = -3*(  ) + c = (a+1) + (b+1) + c = (a + b + c) + 2.


      The sum (a + b + c) is the coefficient of the given polynomial  p(x) = x^3 + 4  at "x"; so, it is 0 (zero, ZERO) :

          a + b + c = 0.


      Therefore,  k = 0 + 2 = 2.     ANSWER


ANSWER.  k = 2.


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It is how the given problem  is assumed to be solved  and how it  should be solved.



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