Trinomial Mystery 2

Let $P_n = a^n + b^n + c^n$ , where $n$ is a positive integer; and $S_1 = a+b+c$ , $S_2 = ab + bc + ca$ , and $S_3 = abc$ . Using Newton's sum or Newton's identities as follows:

$\begin{aligned} P_1 & = S_1 = a+ b+ c = 1 \\ P_2 & = S_1P_1 - 2S_2 = (1)(1) - 2S_2 = 2 & \small \color{#3D99F6} \implies S_2 = - \frac 12 \\ P_3 & = S_1P_2 - S_2P_1 + 3S_3 = (1)(2) + \frac 12 (1) + 3S_3 = 3 & \small \color{#3D99F6} \implies S_3 = - \frac 16 \\ P_4 & = S_1P_3 - S_2P_2 + S_3P_1 = (1)(3) + \frac 12 (2) + \frac 16(1) = \frac {25}6 \\ P_5 & = S_1P_4 - S_2P_3 + S_3P_2 = (1)\left(\frac {25}6 \right) + \frac 12 (3) + \frac 16(2) = \boxed 6 \end{aligned}$

To begin with, let's write down the equations:

$a+b+c=1$ ••• $\boxed{1}$

$a^{2}+b^{2}+c^{2}=2$ ••• $\boxed{2}$

$a^{3}+b^{3}+c^{3}=3$ ••• $\boxed{3}$

Let's approach the problem by trying to reach $a^{5}+b^{5}+c^{5}$ by expanding the multiplication of the $2^{nd}$ and $3^{rd}$ equations, grouping the terms, substituting values from the $1^{st}$ equation, and distributing:

$(a^{2}+b^{2}+c^{2})(a^{3}+b^{3}+c^{3})$

$=a^{5}+a^{2}b^{3}+a^{2}c^{3}+a^{3}b^{2}+b^{5}+b^{2}c^{3}+a^{2}c^{3}+b^{3}c^{2}+c^{5}$

$=a^{5}+b^{5}+c^{5}+a^{2}b^{2}(a+b)+b^{2}c^{2}(b+c)+a^{2}b^{2}(a+c)$

$=a^{5}+b^{5}+c^{5}+a^{2}b^{2}(1-c)+b^{2}c^{2}(1-a)+a^{2}b^{2}(1-b)$

$=\boxed{a^{5}+b^{5}+c^{5}+a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}-abc(ab+bc+ac)=6}$

In an attempt to produce $ab+bc+ac$ , let’s square the $1^{st}$ equation, expand and factor:

$(a+b+c)^{2}=1^{2}$

$a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=1$

Let’s substitute the $2^{nd}$ equation and solve for $ab+bc+ac$ :

$2+2(ab+bc+ac)=1$

Subtract $2$ from both sides:

$2(ab+bc+ac)=-1$

Divide both sides by $2$ :

$\boxed{ab+bc+ac=-\frac{1}{2}}$

To produce $abc$ , let’s raise the $1^{st}$ equation to the $3^{rd}$ power, expand, and factor:

$(a+b+c)^{3}=1^{3}$

$a^{3}+b^{3}+c^{3}+3(a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b))+6abc=1$

Let’s use the $1^{st}$ equation to substitute values:

$a^{3}+b^{3}+c^{3}+3(a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c))+6abc=1$

Let’s distribute:

$a^{3}+b^{3}+c^{3}+3(a^{2}+b^{2}+c^{2}-(a^{3}+b^{3}+c^{3})+6abc=1$

Let’s substitute the $3^{rd}$ equation in:

$3+3(a^{2}+b^{2}+c^{2}-(3))+6abc=1$

Let’s substitute the $2^{nd}$ equation:

$3+3((2)-(3))+6abc=1$

After simplifying,

$6abc=1$

Dividing both sides by $6$ ,

We get $\boxed{abc=\frac{1}{6}}$ .

Let’s produce $a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}$ by squaring this equation we solved earlier, $ab+bc+ac=-\frac{1}{2}$ :

$(ab+bc+ac)^{2}=(-\frac{1}{2})^{2}$

Expanding and factoring,

$a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}+2abc(b+a+c)=\frac{1}{4}$ .

Substituting $abc=\frac{1}{6}$ and the $1^{st}$ equation,

$a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}+2(\frac{1}{6})(1)=\frac{1}{4}$ .

Simplifying,

$a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}+\frac{1}{3}=\frac{1}{4}$ .

Subtracting $\frac{1}{3}$ from both sides,

$a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}=\frac{1}{4}-\frac{1}{3}$ .

Simplifying,

$\boxed{a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}=-\frac{1}{12}}$ .

Substituting the boxed equations for their respective left-hand sides that are in the equation $a^{5}+b^{5}+c^{5}+a^{2}b^{2}+b^{2}c^{2}+a^{2}c^{2}-abc(ab+bc+ac)=6$ ,

$a^{5}+b^{5}+c^{5}+(-\frac{1}{12})-\frac{1}{6}(-\frac{1}{2})=6$ .

Simplifying, $\boxed{\boxed{a^{5}+b^{5}+c^{5}=6}}$ .

The quantities a, b, c are the roots of the equation $x^3$ - $x^2$ - $\dfrac{x}{2}$ - $\dfrac{1}{6}$ =0. Multiplying by x we get $a^4$ + $b^4$ + $c^4$ =coefficient of x in the equation $x^4$ - $x^3$ - $\dfrac{1}{2}$ $x^2$ - $\dfrac{x}{6}$ =0, which is $\dfrac{25}{6}$ . Repeating this process once again, we get $a^5$ + $b^5$ + $c^5$ =6