Question

The equation x^3+x-1=0 has three roots a, b, c
let S(n)= a^n+b^n+c^n
Find the values for S(6) and S(8)

Answer 1

Yep, I know so far, but what comes next?

Answer 2

oh my god! How am I going to memorize formulae that complicated?!

Answer 3

the answer simply said that S6=-2S4-S2+3=1
And S8=-2S6-S4+S2=-6

Answer 4

I'm going to sleep now, so see you tmr! thanks a lot for your help!

Answer 5

The equation x^3+x-1=0 has three roots a, b, c
let S(n)= a^n+b^n+c^n
Find the values for S(6) and S(8)

$$x^3+x-1=0\\a+b+c=0\\ab+bc+ac=1\\abc=1$$so we can infer:$$(a+b+c)^3=0\\a^2+b^2+c^2+2(ab+bc+ac)=0\\a^2+b^2+c^2=-2$$

Answer 6

$$(ab+bc+ac)^2=a^2b^2+b^2c^2+a^2c^2+2(ab^2c+a^2bc+abc^2)\\1=a^2b^2+b^2c^2+a^2c^2+2(a+b+c)=a^2b^2+b^2c^2+a^2c^2\\(a^2+b^2+c^2)^2=a^4+b^4+c^4-2(a^2b^2+b^2c^2+a^2c^2)=a^4+b^4+c^4-2\\4=a^4+b^4+c^4-2\\S(4)=a^4+b^4+c^4=6$$as an example

Answer 7

a similar approach will work for \(S(8)\)

Answer 8

\(S(6)\) can be done by cubing above

Answer 9

is this like competition math? :-)

Answer 10

Ok thanks a lot! Actually is a pasr paper question from Cambridge International Examination Further math

Answer 11

$$x^3+x-1=0\\a+b+c=0\\ab+bc+ac=1\\abc=1$$Okay, consider:$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=-2\\S(2)=-2\\a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+a^2c^2)\\a^2b^2+b^2c^2+a^2c^2=(ab+bc+ac)^2-2(ab^2c+ab^2c+abc^2)\\a^2b^2+b^2c^2+a^2c^2=1-2(a+b+c)=1\\a^4+b^4+c^4=4-2(1)=2\\S(4)=2$$further$$a^8+b^8+c^8=(a^4+b^4+c^4)^2-2(a^4b^4+b^4c^4+a^4c^4)\\a^4b^4+b^4c^4+a^4c^4=(a^2b^2+b^2c^2+a^2c^2)^2-2(a^4b^2c^2+a^2b^4c^2+a^2b^2c^4)\\a^4b^4+b^4c^4+a^4c^4=1-2(a^2b^2c^2)(a^2+b^2+c^2)=1-2(-2)=5\\S(8)=a^8+b^8+c^8=4-2(5)=-6$$

Answer 12

I still do not understand how you get sigma(a^4+b^4)