1A. Let $a$ and $b$ be the roots of the equation $x^{2}-3 c x-8 d=0$, and $c$ and $d$ be the roots of the equation $x^{2}-3 a x-8 b=0$. Calculate the sum $a+b+c+d$, if $a, b, c, d$ are distinct real numbers.
Solution. From Vieta's formulas, we have $a+b=3c$, $c+d=3a$, $ab=-8d$, $cd=-8b$. If we add the first two equations, we get $a+b+c+d=3(a+c)$. From these, we find: $b=3c-a$, $d=3a-c$, and by substituting into the other two equations, we get: $a(3c-a)=-8(3a-c)$, $c(3a-c)=-8(3c-a)$.

If we subtract these two equations, we get: $c^2-a^2=32(c-a)$.

Since $a \neq c$, it follows that $a+c=32$. Then $a+b+c+d=3(a+c)=32 \cdot 3=96$.