Maths(HKMO)

Theorem: If k,m,n are three distinct roots of ax^2+bx+c, then a=b=c=0.
Proof: The quadratic equation has at most 2 distinct roots. If k,m,n are three distinct roots, then it is identically equal to 0. i.e. a=b=c=0

Now multiply the given equation by (a+x)(b+x)(c+x):
a^2(b+x)(c+x) / [(a-b)(a-c)] + b^2 (a+x)(c+x)/ [(b-c)(b-a)] + c^2(a+x)(b+x) / [(c-a)(c-b)] ≡ p+qx+rx^2

Put x=-a,-b,-c respectively.
a^2=p-qa+ra^2-------------(1a)
b^2=p-qb+rb^2-------------(2a)
c^2=p-qc+rc^2-------------(3a)
we can obtain,
a^2(r-1)-qa-p=0------------(1b)
b^2(r-1)-qb-p=0------------(2b)
c^2(r-1)-qc-p=0------------(3b)

So, a,b,c are three distinct roots of (r-1)x^2-qx-p=0
By the above theorem, p=0, q=0,r=1
s=7p+8q+9r=9

The value of s is 9.

No. It is definitely c^2 / [(c-a)(c-b)(b+x)]

2010-01-01 12:54:20 補充：
Thanks for cheukyan4004's answer.
But I want to ask is what is the theorem you mentioned in your answer?

Any typing error of c^2 / [(c-a)(c-b)(b+x)] ?