You can
put this solution on YOUR website! 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!
(