Question

if a+b+c=1
a²+b²+c²=2
a³+b³+c³=3
find a¹⁰+b¹⁰+c¹⁰?​

Answer 1

Step-by-step explanation:

Let a,b,ca,b,c be the roots of x3−px2+qx−r=0…….(1)x3−px2+qx−r=0…….(1)

where p=a+b+c=1p=a+b+c=1 ; q=ab+bc+caq=ab+bc+ca ; r=abcr=abc

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⟹(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⟹

1=2+2q⟹q=−121=2+2q⟹q=−12

Plugging x=a,b,x=a,b, c in (1) and adding

a3+b3+c3−(a2+b2+c2)−12(a+b+c)−3r=0a3+b3+c3−(a2+b2+c2)−12(a+b+c)−3r=0

3–2−12=3r⟹r=163–2−12=3r⟹r=16

So (1)(1) becomes x3−x2−x2−16=0…….(2)x3−x2−x2−16=0…….(2)

Multiplying (2)(2) by xx , x4−x3−x22−x6=0…….(3)x4−x3−x22−x6=0…….(3)

As before

a4+b4+c4−(a3+b3+c3)−12(a2+b2+c2)−a+b+c6=0a4+b4+c4−(a3+b3+c3)−12(a2+b2+c2)−a+b+c6=0

a4+b4+c4−3−12∗2−16=0a4+b4+c4−3−12∗2−16=0

a4+b4+c4=256a4+b4+c4=256