2
$\begingroup$

Question : $$a+b+c = 0\\ a^3 +b^3 +c^3 = 12\\ a^5 +b^5 +c^5 = 40$$ Then , $a^4 + b^4 + c^4 = ?$

My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think $3abc = 12$ and got $abc = 4$ After that I am unable to link it with any other equation

$\endgroup$
  • $\begingroup$ This question has been asked already here. It is also popular on the web in general, see here. $\endgroup$ – Dietrich Burde Oct 23 '18 at 18:09
  • $\begingroup$ Link please , thanks for informing $\endgroup$ – user580093 Oct 23 '18 at 18:10
  • $\begingroup$ Looking for links (method is always the same) on this site, see here. $\endgroup$ – Dietrich Burde Oct 23 '18 at 18:12
  • $\begingroup$ $\displaystyle a^4+b^4+c^4=8.$ $\endgroup$ – jacky Oct 23 '18 at 18:55
6
$\begingroup$

Denote $S_n = a^n + b^n + c^n$. So we have $S_0 = 3, S_1 = 0, S_3 = 12$ and $S_5 = 40$.

Consider $f(x) = (x-a)(x-b)(x-c)$. Write $f(x) = x^3 - ux^2 - vx - w$. Then it is easy to see $u = 0$ and $v = -(ab + bc + ca) = \frac{S_2}2$. This means $a^3 = va + w$, so $a^n = va^{n-2} + wa^{n-3}$, and do the same on $b, c$ and summing together gives you $S_n = vS_{n-2} + wS_{n-3}$.

This should give you enough information set up a system of equations to see $S_4 = 8$. For instance, the first thing you can do is

$S_3 = vS_1 + wS_0$, so $12 = 3w$ and $w = 4$.

$\endgroup$
2
$\begingroup$

Hint: $$(a+b+c)^4=a^4+b^4+c^4+4a^3(b+c)+4b^3(a+c)+4c^3(a+b)+6(a^2b^2+b^2c^2+c^2a^2)+12abc(a+b+c)=0$$

$\endgroup$
  • $\begingroup$ $$ab+bc+ac=-\frac{1}{2}(a^2+b^2+c^2)$$ and squaring. $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '18 at 18:24
  • $\begingroup$ Sorry , sir but after putting values my whole equation becomes zero, I am also doing correct $\endgroup$ – user580093 Oct 23 '18 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.