The key to do this without getting tangled up is using symmetry – the fact that all your answers have to look the same if you swap round α,β,γ.
The cubic equation x3+2x2−3x−5 = 0 has roots α,β,γ.
You want to make a new equation with roots α+β,β+γ,γ+α, without solving your original equation.
You can calculate from the original equation
α+β+γ = −2
αβ+βγ+γα = −3
αβγ = 5
(If the original equation had a coefficient of x3 different from 1, simplest to divide the whole thing through to get that coefficient equal to 1 and the equation in a standard form).
You want the sum of roots, the sum of products-in-pairs of roots, and the product of roots of the new equation.
Sum of roots of the new equation is easy;
(α+β)+(β+γ)+(γ+α) = 2(α+β+γ)
So, what’s the sum of products-in-pairs?
Look through the expression, and you see you have 3αβ in the result (one αβ from each multiplication). Since it’s symmetrical, you must have 3βγ and 3γα too.
Look through again, and you see you have one β2 in the result (from the first multiplication only). Since it’s symmetrical, you must have one α2 and one γ2 too.
So the sum of products in pairs of roots of the new equation is (α2+β2+γ2) + 3(αβ+βγ+γα). We’ve got this without a lot of calculation – by looking carefully and using symmetry.
Not quite enough, though, because we don’t know off-hand what α2+β2+γ2 is.
We do know what (α+β+γ)2 is.
And the same looking and thinking about symmetry tells us (α+β+γ)2 = α2+β2+γ2 + 2(αβ+βγ+γα)
Now we know that the the sum of products in pairs of roots of the new equation is (α+β+γ)2 +(αβ+βγ+γα).
What about the product of all three roots of the new equation?
Look through the multiplication. One α2.β. Two αβγ.
Using symmetry, the answer is
α2.β+β2.γ+γ2.α+α2.γ+β2.α+γ2.β + 2αβγ
Getting this in shape to use what we know
(α+β+γ)·(αβ+βγ+γα) = α2.β+β2.γ+γ2.α+α2.γ+β2.α+γ2.β + 3αβγ
(Again, just look to see how many α2.β and how many αβγ, and use symmetry)
(α+β)·(β+γ)·(γ+α) = (α+β+γ)·(αβ+βγ+γα)−αβγ
Here’s a video about this sort of problem (not brilliant, but it’ll do)