#alpha, beta# are real numbers such that #alpha^3-3alpha^2+5 alpha - 17=0# and #beta^3-3beta^2+5beta+11 = 0#. What is the value of #alpha+beta# ?

First use Tschirnhaus transformations to simplify the cubics:

#0 = alpha^3-3alpha^2+5alpha-17 = gamma^3+2gamma-14#

#0 = beta^3-3beta^2+5beta+11 = delta^3+2delta+14#

where #gamma = alpha-1# and #delta = beta-1#

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

We can check the discriminant of these cubics, finding that they are both negative, so they have #1# Real root and #2# Complex roots each.

By Descartes' rule of signs, the Real roots result in #gamma > 0# and #delta < 0#. We use this later.

Adding these two equations, we get:

#0 = gamma^3+delta^3+2gamma+2delta#

#color(white)(0) = (gamma+delta)(gamma^2-gamma delta+delta^2)+2(gamma+delta)#

#color(white)(0) = (gamma+delta)(gamma^2-gamma delta+delta^2+2)#

So either #gamma+delta = 0# or #gamma^2-gamma delta+delta^2+2 = 0#

Note however that #gamma*delta < 0#, hence #gamma^2-gamma delta+delta^2 + 2 > 0# for any Real values of #gamma# and #delta#.

So the only Real solution gives us #gamma+delta = 0# and hence #alpha+beta = 2#