Cubes and Sixths

The Key technique is Newton's Sum .

At first notice that $\frac{a^6-1}{a-1} = \frac{(a-1)(a^5+a^4+a^3+a^2+a+1)}{a-1)} = a^5+a^4+a^3+a^2+a+1$

Similarly, $\frac{b^6-1}{b-1} = b^5+b^4+b^3+b^2+b+1$

and $\frac{c^6-1}{c-1} = c^5+c^4+c^3+c^2+c+1$ .

Now, we know that if the roots of some polynomial $P(x)= a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ were $r_1,r_2,r_3,...r_{n-1},r_{n}$ and $P_k =\displaystyle \sum_{i=1}^n r_i^k$ then the following equations would be true.

$1) a_nP_1 + a_{n-1} =0$

$2) a_nP_2 + a_{n-1}P_1 + 2 a_{n-2} = 0$

$3) a_nP_3 + a_{n-1}P_2 + a_{n-2}P_1 + 3a_{n-3} =0$

$4) a_nP_4 + a_{n-1}P_3 + a_{n-2}P_2 + a_{n-3}P_1 + 4a_{n-4} =0$

$5) a_nP_5 + a_{n-1}P_4 + a_{n-2}P_3 + a_{n-3}P_2 + 4a_{n-4}P_1 + 5a_{n-5} =0$

Now consider the polynomial $g(x) = x^5 + 3x^3 +5x^2$ .

Notice that it's roots are $a,b,c,0$ and $0$ .

So, in this case $P_k = a^k+b^k+c^k+0^k+0^k = a^k+b^k+c^k$ .

Now, by using The equations $1,2,3,4,5$ we can easily see that for $g(x)$ ,

$P_1 = 0$ ,

$P_2 = -6$ ,

$P_3 = -15$ ,

$P_4 = 18$

and $P_5 =75$ .

From here the rest is pretty easy.

Since, $\frac{a^6-1}{a-1} + \frac{b^6-1}{b-1} +\frac{c^6}{c-1}$

$= a^5+a^4+a^3+a^2+a+1+b^5+b^4+b^3+b^2+b+1 +c^5+c^4+c^3+c^2+c+1$ .

$=P_5+P_4+P_3+P_2+P_1+3$

$=75$

Note that : $a^6-1=(3a+5)^2-1=9a^2+30a+24$ , so : $\frac{a^6-1}{a-1} = \frac{9a^2+30 a +24 }{a-1} = 39+9a + \frac{63}{a-1}$ Now : $S= \sum_{cyc} \frac{a^6-1}{a-1} = \sum_{cyc} 39 +9a +\frac{63}{a-1}$ $S=117 + 9(a+b+c) + 63 \cdot \frac{ 3+(ab+bc+ac)-2(a+b+c)}{(a+b+c)+(abc)-(ab+ac+bc)-1}$ From Viete formula we get : $a+b+c=0;ab+bc+ac=3;abc=-5$ , apply to get : $S= 117 + 63 \cdot \frac{3+3}{-5-3-1} = 117 - 42 =75$

Because $\frac{a^6-1}{a-1}=a^5+a^4+a^3+a^2+a+1$ and because $a^3+3a+5=0$ , we can write $0=(a^2+a-2)( a^3+3a+5)=a^5+a^4+a^3+8a^2-a-10$ Subtracting this from the 1st equation gives $\frac{a^6-1}{a-1}=-7a^2+2a+11$ Similarly for b and c, so $\frac{a^6-1}{a-1}+ \frac{b^6-1}{b-1}+ \frac{c^6-1}{c-1}=-7(a^2+b^2+c^2)+2(a+b+c)+33$

We have $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=x^3+3x+5$ so $a+b+c=0$ $ab+bc+ca=3$ $-abc=5$

Because $(a+b+c)^2= a^2+b^2+c^2+2(ab+bc+ca)=a^2+b^2+c^2+6=0$ we have $a^2+b^2+c^2=-6$

$\frac{a^6-1}{a-1}+ \frac{b^6-1}{b-1}+ \frac{c^6-1}{c-1}=-7(-6)+2(0)+33=\boxed{75}$