a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

nhân 2 vế cho (a+b+c) ta được:

a+b+c/a+b   +  a+b+c/b+c   +   a+b+c/c+a= a+b+c/90

1 + c/a+b + 1+ a/b+c + 1+ b/c+a=2007/90

c/a+b + a/b+c + b/c+a= 2007/90 - 3=? tự tính

vậy kết quả cần tìm là: 

a)        \(a+\frac{1}{a}=3\)

\(\Leftrightarrow\)\(\left(a+\frac{1}{a}\right)^2=9\)

\(\Leftrightarrow\)\(a^2+2+\frac{1}{a^2}=9\)

\(\Leftrightarrow\)\(a^2+\frac{1}{a^2}=7\)

  Ta có:      \(\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right)=3.7\)

\(\Leftrightarrow\)\(a^3+\frac{1}{a}+a+\frac{1}{a^3}=21\)

\(\Leftrightarrow\)\(a^3+\frac{1}{a^3}=21-3=18\)

Ta lại có:    \(\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)=7.18\)

\(\Leftrightarrow\)\(a^5+\frac{1}{a}+a+\frac{1}{a^5}=126\)

\(\Leftrightarrow\)\(a^5+\frac{1}{a^5}=126-3=123\)