# Solve this question

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

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

## 2 Answers

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 $$w = 4$$.

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

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