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

Algebra ->  Test -> 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      Log On


   

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) About Me  (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(52780) About Me  (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)  1%2F%28a%5E2+-+a+%2B+1%29 = %28a%2B1%29%2F%28a%5E3%2B1%29.


     It follows from the standard identity  a%5E3%2B1 = %28a%2B1%29%2A%28a%5E2-a%2B1%29.


     Further, since "a" is the root of the equation  x%5E3%2B4 = 0,  we have  


         a%5E3%2B4 = 0;  hence,   a%5E3%2B1 = %28a%5E3%2B4%29-3 = 0-3 = -3.   


     Therefore,  1%2F%28a%5E2+-+a+%2B+1%29 = %28a%2B1%29%2F%28a%5E3%2B1%29 = -%28a%2B1%29%2F3.    (1)



(2)  Similarly,  1%2F%28b%5E2+-+b+%2B+1%29 = -%28b%2B1%29%2F3.             (2)


(3)  Therefore,


         k = - 3[(1/(a² - a + 1) + (1/(b² - b + 1)] + c  = -3*( -%28a%2B1%29%2F3+-+%28b%2B1%29%2F3 ) + 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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