6. Let $a, b, c$ be nonnegative rational numbers. If $a(b+c)=36, b(a+c)=50$ and $c(a+b)=56$, what is $a b c$?
First solution.

Given are the equations

$$
\begin{aligned}
a b + a c & = 36 \\
a b + b c & = 50 \\
a c + b c & = 56
\end{aligned}
$$

By adding the previous three equations, we get:

$$
\begin{gathered}
a b + a c + a b + b c + a c + b c = 142 \\
2(a b + a c + b c) = 142 \\
a b + a c + b c = 71
\end{gathered}
$$

By subtracting the given equations from the last equation, we get:

$$
\begin{gathered}
b c = 71 - 36 \\
b c = 35
\end{gathered}
$$

$$
a c = 71 - 50
$$

$$
a c = 21
$$

$$
\begin{gathered}
a b = 71 - 56 \\
a b = 15
\end{gathered}
$$

The product of the obtained equations

$$
a b \cdot a c \cdot b c = 15 \cdot 21 \cdot 35
$$

is equal to

$$
\begin{gathered}
a b \cdot a c \cdot b c = 3 \cdot 5 \cdot 3 \cdot 7 \cdot 5 \cdot 7 \\
a b c \cdot a b c = (3 \cdot 5 \cdot 7) \cdot (3 \cdot 5 \cdot 7) \\
a b c \cdot a b c = 105 \cdot 105
\end{gathered}
$$

Since $a, b, c$ are non-negative rational numbers, $a b c$ cannot be -105, so we conclude that $a b c = 105$.

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