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$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3} \text{then}\ a^5+b^5+c^5= \ ?$$

A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.

Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.

What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.

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Let's start with the basic symmetric expressions: $ab+bc+ca$ and $abc$. You can refer to giannispapav's answer for details, which shows that $$ab+bc+ca = -1/2, abc = 1/6.$$

With that, Vieta's formulas implies that $a,b,c$ satisfy: $$ x^3 -x^2 - x/2 -1/6=0,\tag{1}$$ Or $$x^3 = x^2 + x/2 + 1/6.$$

That means, for $x$ equals $a,b,c$, $$x^4 = x^3 + x^2/2 + x/6,$$ and $$x^5 = x^4 + x^3/2 + x^2/6.$$ Adding the two equations above, we have $$x^5 = \frac32x^3 + \frac23x^2 + \frac16x.$$ Now replace $x$ as $a,b,c$ and add them all up, we have $$a^5+b^5+c^5 = \frac32(a^3+b^3+c^3) + \frac23(a^2+b^2+c^2) + \frac16(a+b+c).$$


Note: if you feels that $$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$$ is too complicated to verify, then Vieta's formulas is the way to go. That is, replace $a,b, c$ in Equation $1$ and add them up, where $1/6$ is indeed $abc$ as in Vieta's formula.

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Using Newton's identities

$$ \begin{aligned} e_{1}&=p_{1}\\ 2e_{2}&=e_{1}p_{1}-p_{2}\\ 3e_{3}&=e_{2}p_{1}-e_{1}p_{2}+p_{3}\\ 4e_{4}&=e_{3}p_{1}-e_{2}p_{2}+e_{1}p_{3}-p_{4}\\ 5e_{5}&=e_{4}p_{1}-e_{3}p_{2}+e_{2}p_{3}-e_{1}p_{4}+p_{5}\\ \end{aligned} $$ with $p_1=1,p_2=2,p_3=3,e_4=0, e_5=0$, we get $p_5 = 6$.

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This answer is almost in the same spirit as @Quang Hoang's, but I hope this answer will add something. Let $$ P(z) = (z-a)(z-b)(z-c)=z^3-\sigma_1 z^2+\sigma_2 z-\sigma_3 $$ where $\sigma_1=a+b+c$, $\sigma_2=ab+bc+ca$ and $\sigma_3=abc$ by Vieta's formula. Note that for $z\in \{a,b,c\}$, $$ z^{n+3} =\sigma_1 z^{n+2}-\sigma_2 z^{n+1}+\sigma_3 z^n, $$ hence by summing over $z\in \{a,b,c\}$, we get recurrence relation $$ s_{n+3}= \sigma_1 s_{n+2}-\sigma_2 s_{n+1}+\sigma_3 s_n $$ for $s_n = a^n+b^n+c^n$. Given the data, it can be easily noted that $$\sigma_1=1 ,\quad \sigma_2 =\frac 12 \left((a+b+c)^2-(a^2+b^2+c^2)\right)=-\frac 12.$$ And by plugging $n=0$, we obtain $$ 3=1\cdot 2+\frac 12\cdot 1 +\sigma_3 s_0=2.5 + \sigma_3s_0, $$ so $\sigma_3=abc\ne 0$ and $s_0=a^0+b^0+c^0=3$. This gives $\sigma_3=\frac 1 6$, implying that $$ s_{n+3}=s_{n+2}+\frac 12 s_{n+1}+\frac 1 6 s_{n},\quad \forall n\ge 0. $$ Now $s_4 =\frac {25}{6}$ and $s_5=6$ follows from the initial data $(s_3,s_2,s_1)=(3,2,1)$.
Note : The theory of homogeneous linear difference equations is behind it.

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You can use

$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)$,

$(ab+ac+bc)^2=a^2+b^2+c^2+2(ab+ac+bc)$,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$,

$(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$

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  • $\begingroup$ How could one guess such relations? $(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$? $\endgroup$ – Paracosmiste Apr 10 at 14:46
  • $\begingroup$ @BPP I don't really know $\endgroup$ – giannispapav Apr 10 at 15:47
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Fun video!

Much time was spent on finding $abc=1/6$.

Alternative method for this: $$\begin{align}a^2+b^2&=2-c^2 \Rightarrow \\ (a+b)^2-2ab&=2-c^2 \Rightarrow \\ (1-c)^2-2ab&=2-c^2 \Rightarrow \\ ab&=c^2-c-\frac12 \Rightarrow \\ abc&=c^3-c^2-\frac c2 \end{align}$$ Similarly: $$abc=a^3-a^2-\frac a2\\ abc=b^3-b^2-\frac b2$$ Now adding them up: $$3abc=(a^3+b^3+c^3)-(a^2+b^2+c^2)-\frac12(a+b+c)=3-2-\frac12 \Rightarrow abc=\frac16.$$ In fact, you can find other terms as well: $$ab+bc+ca=(a^2+b^2+c^2)-(a+b+c)-\frac32=2-1-\frac32=-\frac12;\\ a^2b^2+b^2c^2+c^2a^2=ab(c^2-c-\frac12)+bc(a^2-a-\frac12)+ca(b^2-b-\frac12)=\\ abc(a+b+c)-3abc-\frac12(ab+bc+ca)=\\ \frac16-\frac12+\frac14=-\frac1{12}$$ Hence: $$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\ 2\cdot 3-(-\frac1{12})+\frac16\cdot (-\frac12)=6.$$

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I think to use an identity must be the shortest way:-

${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)} $

If we square the 1st equation we get $ {ab+bc+ca=\frac {-1}2} $

On squaring the above equation we get $ {(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\frac 14} $

We also know ${ (a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca) }$ from here we get :- $abc=\frac 16$

Substituting the known values We get :- ${ (ab)^2+(bc)^2+(ca)^2=\frac {-1}{12} } $

Now squaring the 2nd equation:- ${ (a^4+b^4+c^4)+2\big ((ab)^2+(bc)^2+(ca)^2)\big )=4 }$

From here we get $(a^4+b^4+c^4)=\frac{25}6$

Now we have all the values to be substituted in the identity that i mentioned therefore you get

$ { (a^5+b^5+c^5)= (\frac{25}{6}\cdot 1) - (3 \cdot \frac {-1}2)+ (\frac 16 \cdot 2) }$

$\Rightarrow$$(a^5+b^5+c^5)=6$

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