Question 10
From x^3 +3x +9 = 0 where roots are a,b,c
First find the equation where roots are a^3, b^3, c^3:
So sub x^(1/3) for x in the original equation gives:
x + 3x^(1/3) + 9 =0
x+9 = -3x^(1/3)
Cubing both sides gives:
x^3 + 27x^2 +243x + 729 = -27x
x^3 + 27x^2 +270x + 729 = 0 <---- equation where roots are a^3, b^3, c^3
So thus a^6 + b^6 + c^6 = (a^3+ b^3+ c^3)^2 - 2(a^3b^3+ a^3c^3+c^3b^3)
= (-27)^2 - 2(270)
= 189
So sub a^3, b^3, c^3 into x^3 + 27x^2 +270x + 729 = 0 gives
a^9 + 27a^6 +270a^3 + 729 = 0
b^9 + 27b^6 +270b^3 + 729 = 0
c^9 + 27c^6 +270c^3 + 729 = 0
Adding them up gives
a^9 + b^9 +c^9 +27(a^6+b^6+c^6) +270(a^3+b^3+c^3) +3*729 = 0
thus a^9 + b^9 +c^9 +27(189) +270(-27) +2187= 0
Thus a^9 +b^9 +c^9 = 0
sorry about the crappy formating... there is prob an easier way lol.