# Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$

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

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.

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.

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 $$p_5 = 6$$.

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.

You can use

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

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

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

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

• How could one guess such relations? $(a+b+c)^5-a^5-b^5-c^5=(a+b)(a+c)(b+c)(a^2+b^2+c^2+ab+ac+bc)$? – user5402 Apr 10 '19 at 14:46
• @BPP I don't really know – 1123581321 Apr 10 '19 at 15:47

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: $$a^5+b^5+c^5=(a^2+b^2+c^2)(a^3+b^3+c^3)-(a^2b^2+b^2c^2+c^2a^2)+abc(ab+bc+ca)=\\ 2\cdot 3-(-\frac1{12})+\frac16\cdot (-\frac12)=6.$$

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

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

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