question for pure maths (20pt)

As a, b and c are the roots of x^3 - 2x^2 + 5x - 3 = 0, then we have

a^3 - 2a^2 + 5a - 3 = 0 ..... (1)
b^3 - 2b^2 + 5b - 3 = 0 ..... (2)
c^3 - 2c^2 + 5c - 3 = 0 ..... (3).

(1) + (2) + (3),
a^3 + b^3 + c^3 - 2(a^2 + b^2 + c^2) + 5(a + b + c) - 3(3) = 0

Let S1 = a + b + c, S2 = a^2 + b^2 + c^2, S3 = a^3 + b^3 + c^3,
we have S3 - 2(S2) + 5(S1) - 3(3) = 0.
I think you can find S1 and S2 yourself and I don't do it for you here.

Similarly, we have
a(a^3 - 2a^2 + 5a - 3) = 0
b(b^3 - 2b^2 + 5b - 3) = 0
c(c^3 - 2c^2 + 5c - 3) = 0.

a^4 - 2a^3 + 5a^2 - 3a = 0 ..... (4)
b^4 - 2b^3 + 5b^2 - 3b = 0 ..... (5)
c^4 - 2c^3 + 5c^2 - 3c = 0 ..... (6).

(4) + (5) + (6),
a^4 + b^4 + c^4 - 2(a^3 + b^3 + c^3) + 5(a^2 + b^2 + c^2) - 3(a + b + c) = 0

Let S4 = a^4 + b^4 + c^4,
we have S4 - 2(S3) + 5(S2) - 3(S1) = 0.

As you can see, the relationship will become S(n+3) - 2[S(n+2)] + 5[S(n+1)] - 3(Sn) = 0, where n is any positive integer greater than or equal to 1.

a^3+b^3+c^3 =(a+b+c)^3 - 3(ab+bc+ac)(a+b+c)+3abc
a^4+b^4+d^4 =(a+b+c)^4 - 4(a+b+c)^2(ab+bc+ac)+4abc(a+b+c)+2(ab+bc+ac)^2
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ｓｏｌｕｔｉｏｎ：
a+b+c=2
ab+bc+ac=5
abc=3

a^3+b^3+c^3 =(a+b+c)^3 - 3(ab+bc+ac)(a+b+c)+3abc
=(2)^3 - 3(5)(2)+3(3)
= -13

a^4+b^4+d^4 =(a+b+c)^4 - 4(a+b+c)^2(ab+bc+ac)+4abc(a+b+c)+2(ab+bc+ac)^2
=(2)^4 - 4(2)^2(5)+4(3)(2)+2(5)^2
=10

2007-02-13 09:13:43 補充：
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