34. If $a+x^{2}=2015, b+x^{2}=2016, c+x^{2}=2017$, and $a b c=24$, then the value of $\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c}$ is . $\qquad$
Answer: $\frac{1}{8}$
Solution: From $a+x^{2}=2015, b+x^{2}=2016, c+x^{2}=2017$, we know
$$
b-a=1, c-b=1,
$$

Therefore,
$$
a=b-1, c=b+1 .
$$

So
$$
a b c=(b-1) b(b+1)=24 \text {, }
$$

which means
$$
b^{3}-b-24=0,
$$

The left side of the equation can be factored, resulting in

Since
$$
b^{2}+3 b+8=\left(b+\frac{3}{2}\right)^{2}+\frac{23}{4}>0 \text {, }
$$

Therefore, $b=3, a=2, c=4$.

Thus
$$
\begin{aligned}
& \frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}-\frac{1}{a}-\frac{1}{b}-\frac{1}{c} \\
= & \frac{2}{12}+\frac{3}{8}+\frac{4}{6}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4} \\
= & \frac{1}{8}
\end{aligned}
$$