Real Root Sum

Factorize ${a}^{3}-15{a}^{2}+20a-50=0$ and $8{b}^{3}-60{b}^{2}-290b+2575$ . ${(a-5)}^{3}-55(a-5)-200=0$ ${(2b-5)}^{3}-220(2b-5)+1600=0$

Let $u = a-5$ and $v = 2b-5$ ; hence $a+b= \frac{2u+v+15}{2}$ .

Now we get ${u}^{3}-55u-200=0$ and ${v}^{3}-220v+1600=0$ .

Multiply the first equation by 8 and subtract the second equation to eliminate the constant: ${8u}^{3}+{v}^{3}-440u-220v=0.$

Factorize the above expression: $(2u+v)[{(2u+v)}^{2}-6uv-220] =0.$

We conclude that $2u=-v$ and substitute this solution to $a+b= \frac{2u+v+15}{2}$ .

Let $a+b=h$ . Then replace $a$ by $h-b$ in the first equation and compare the coefficient of $b^2$ (after proper scaling) with the second equation to get $h$ . The answer turns out to be $7.5$ after a straight forward calculation.