About finding general an+bn+cn,an+bn+cn, there is an elementary aspect to this, although not necessarily what you want. You already know that a+b+c=0,a+b+c=0, and the other answer gives enough to find a2+b2+c2a2+b2+c2 and a3+b3+c3.a3+b3+c3. Call those x1,x2,x3,x1,x2,x3, then solve in perpetuity with
xn+3=xn+1+xnxn+3=xn+1+xn
Two errors in the question,
bc+ca+ab=−1,bc+ca+ab=−1,
abc=1.abc=1.
So far, I get
x1=0,x1=0,
x2=2,x2=2,
x3=3,x3=3,
x4=2,x4=2,
x5=5,x5=5,
x6=5,x6=5,
x7=7,x7=7,
x8=10,x8=10,
x9=12,x9=12,
x10=17,x10=17,
Since the real root of x3−x−1x3−x−1 is slightly larger than 1,1, the numbers increase, are positive and so on. The two complex roots are smaller than 11 in modulus, so, for large n,n, we get xn≈Rnxn≈Rn where R≈1.3247R≈1.3247 is the real root. For example, R10≈16.643
For more in debt answer just ask (hopefully its debt enough through)
-wolfie
