2021/042) For real numbers a and b that satisfy $a^3+12a^2+49a + 69=0$ and $b^3 -9b^2+ 28b -31 = 0$ find a+b.

We have
 $a^3+12a^2+49a + 69=0$
=> $(a+4)^3 + (a+5) = 0$
or $(a+4)^3 + (a+4) + 1= 0\cdots(1)$

$b^3 -9b^2+ 28b -31 = 0$
$=>(b-3)^3 + (b-4) = 0$
$=>(b-3)^3 + (b-3) -1 = 0\cdots(2)$

if we define $f(x) = x^3+x$ knowing that f(x) is odd function
$(f(a+4) = -1 $ and $-f(b-3) = 1$

so $(a+4) = - (b- 3)$

or $a + b = - 1$