The real numbers $\alpha ,\beta$ satisfy the equation ${\alpha }^3-3{\alpha }^2+5\alpha -17=0; {\beta }^3-3{\beta }^2+5\beta +11=0.$ Then $\alpha +\beta$ is equal to

Average value of –17 and 11 is –3

Let $f\left(x\right)=x^3-3x^2+5x-3\Rightarrow {(x-1)}^3+2(x-1)$

The graph of f(x) is odd w.r.t to x = 1

Then if the root of $x^3-3x^2+5x-17=0$ is $\alpha =1+x_1$

Then the root of $x^3-3x^2+5x+11=0$ is $\beta =1-x_1$

So, $\alpha +\beta =2$