## Task B-1.3.

Let $a, b, c$ and $d$ be real numbers different from zero such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0$ and $a+b+c+d+a b c d=0$. Calculate $\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right)$.
## First Solution.

The desired expression is equal to

$$
\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right)=\frac{c d-a b}{a b c d} \cdot \frac{d+c}{c d}=\frac{c d^{2}+c^{2} d-a b d-a b c}{a b c^{2} d^{2}}
$$

From $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=0$ we get $b c d+a c d+a b d+a b c=0$,

which implies $b c d+a c d=-a b d-a b c$.

1 point

Therefore,

$$
\begin{aligned}
\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right) & =\frac{c d^{2}+c^{2} d-a b d-a b c}{a b c^{2} d^{2}} \\
& =\frac{c d^{2}+c^{2} d+b c d+a c d}{a b c^{2} d^{2}}=\frac{c d(d+c+b+a)}{a b c^{2} d^{2}} \\
& =\frac{d+c+b+a}{a b c d}
\end{aligned}
$$

Since from $a+b+c+d+a b c d=0$ it follows that $a+b+c+d=-a b c d$,

we get

$$
\left(\frac{1}{a b}-\frac{1}{c d}\right)\left(\frac{1}{c}+\frac{1}{d}\right)=\frac{d+c+b+a}{a b c d}=\frac{-a b c d}{a b c d}=-1
$$

1 point