If $a+b=8$, $b+c=-3$, and $a+c= -5$, what is the value of the product $abc$?
Let's solve for $b$ in the equation $b+c=-3$. If we subtract $c$ from both sides of the equation, we get $b=-3-c$. Now, let's substitute this expression for $b$ into the equation $a+b=8$. Then we have $a-3-c=8$. So $a-c=11$. Now, let's add this equation to the equation $a+c=-5$. Adding these two equations, we get $2a=6$, which means $a=3$. Since we know $a=3$, we can substitute this into our equation $a-c=11$ to get $3-c=11$. That means $c=-8$. And since we know $a=3$ and $c=-8$, we can find $b$ by substituting into the equation $a+b=8$. We get $3+b=8$, so $b=5$. Now that we know $a=3$, $b=5$, and $c=-8$, we can find the product $abc$. $abc=3*5*-8=-120$.

