Question 1195
 This question is not easy for high school students,since some special
 polynomial identities should be used.

 Sol:
 First identity about the square: a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) ...(*) 
 Second identity about the cube: a^3 + b^3 + c^3 = 
 (a+b+c) ((a+b+c)^2 - 3(ab+bc+ca)) + 3abc ...(**)
 Third identity about the 5th power: a^5 + b^5 + c^5 = 
(a^3 + b^3 + c^3)(a^2 + b^2 + c^2)-(a +b+c)[(ab+bc+ca)^2 -2 abc(a + b+ c)] 
+ abc(ab+ bc+ ca)...(***)

 {The detailed proof of the 2nd & 3rd identities is on the bottom of
  the solution]
  Now,a, b and c are the roots of x^3 - 2x^2 + x +3 = 0 .
 By the relations of roots and coefficients,
 we have a+b+c = 2, ab+bc+ca = 1 ,abc=-3.
 
 From the identity (**),we obtain 
 a^3 + b^3 + c^3 = (a+b+c) ((a+b+c)^2 - 3(ab+bc+ca)) + 3abc 
                 = 2[4  -3] + 3(-3)
                 = -7

 Then by the identity (***) & (*),we obtain  a^5 + b^5 + c^5 = 
 (a^3 + b^3 + c^3) (a^2 + b^2 + c^2)-(a+b+c)[(ab+bc+ca)^2 -2 abc(a + b+ c)] +   abc(ab+ bc+ ca) 
 = (-7)(4-2) - 2[1- 2(-3)*2] + (-3)*1
 = -14 -2 *13 -3
 = -43

 Proof of(**) & (***):
a^3 + b^3 + c^3 -3 abc = a^3 + (b + c)^3 -3bc(b+c)- 3abc 
                        = (a+b+c) (a^2 +(b+c)^2 - a(b+c)) -3bc(a +b+c)
                        = (a+b+c) (a^2 +(b+c)^2 - a(b+c) -3bc)
                        = (a+b+c) (a^2 + b^2+ c^2 - ab -bc-ca )
  [ Use a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca) ...(*) ]
                        = (a+b+c) ((a+b+c)^2 - 3(ab+bc+ca))

 We get a^3 + b^3 + c^3 = (a+b+c) ((a+b+c)^2 - 3(ab+bc+ca)) + 3abc ...(**)

 (a^3 + b^3 + c^3) (a^2 + b^2 + c^2) = a^5 + b^5 + c^5 + a^2b^2(a + b)+ b^2c^2(b + c)+ c^2a^2(c + a)
 = a^5 + b^5 + c^5 + a^2b^2(a +b+c)+ b^2c^2(a+ b + c)+ c^2a^2(a+b+c) - abc(ab+ bc+ ca)
 = a^5 + b^5 + c^5 + a^2b^2(a +b+c)+ b^2c^2(a+ b + c)+ c^2a^2(a+b+c) - abc(ab+ bc+ ca) 
 = a^5 + b^5 + c^5 + (a +b+c)(a^2b^2+ b^2c^2+ c^2a^2) - abc(ab+ bc+ ca)
= a^5 + b^5 + c^5 + (a +b+c)[(ab+bc+ca)^2 -2 abc(a + b+ c)] - abc(ab+ bc+ ca)

 Hence,a^5 + b^5 + c^5 = 
(a^3 + b^3 + c^3) (a^2 + b^2 + c^2)-(a +b+c)[(ab+bc+ca)^2 -2 abc(a + b+ c)] + abc(ab+ bc+ ca)...(***)


 Try to read carefully for every step. Good luck!