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Let $p(x) = x^4 + x + 1 = 0$, and let $a$, $b$, $c$, $d$ be its roots. Find $a^4 + b^4 + c^4 + d^4$.

I have no idea how to start solving this problem.

Carl Mummert
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Md Masood
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    What are you studying? What text is this drawn from, if any? If not, how did the question arise? What kind of approaches (to similar problems) are you familiar with? What kind of answer are you looking for? Basic approach, hint, explanation, something else? Is this question something you think should be able to answer? Why or why not? – Shaun Aug 14 '18 at 20:54
  • This question was related to polynomials. I was looking for a hint. – Md Masood Aug 15 '18 at 13:01

2 Answers2

9

For any of the roots, $r^4=-r-1$ so that the requested sum is $-a-b-c-d-4$, and by the Vieta's formula, the sum of the roots is $0$.

2

As an alternative, let consider

$$(x-a)(x-b)(x-c)(x-d)=$$ $$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$

then

  • $S_1=a+b+c+d=0$
  • $S_2=ab+ac+ad+bc+bd+cd=0$
  • $S_3=abc+abd+acd+bcd=-1$
  • $S_4=abcd=1$

and by Newton's sums we have that

  • $P_1=a+b+c+d=S_1=0$
  • $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2=0$
  • $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3=-3$
  • $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4=-4$
user
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