Polynomial Equation... if α,β,γ are positive real roots of the equation x^3-mx^2+12x-4n=0 such that (α+β )(β+γ)(γ+α)=8αβγ find (m+n)?

α,β,γ are roots of x^3 - mx^2 + 12x - 4n = 0 , so it must be (x - α)(x - β)(x - γ) = 0 .
Therefore , x^3 - mx^2 + 12x - 4n = x^3 - (α + β + γ)x^2 + (αβ + βγ + αγ)x - αβγ .
Thus , we find
α + β + γ = m ---(#1)
αβ + βγ + αγ = 12 ---(#2)
αβγ = 4n ---(#3)
And the condition is
(α + β)(β + γ)(γ + α) = 8αβγ
With (#1) , it becomes
(m - γ)(m - α)(m - β) = 8αβγ
m^3 - (α + β + γ)m^2 + (αβ + βγ + αγ)m - αβγ = 8αβγ
Sumstitute (#1) , (#2) , (#3) into this ,
m^3 - m*m^2 + 12m - 4n = 32n
12m = 36n
m = 3n ---(#4)

From (#2) ,
αβ + (α + β)γ = 12
(α + β)γ = 12 - αβ
γ = (12 - αβ) / (α + β) ---(#5)

Substitute (#5) into (#1) ,
α + β + (12 - αβ) / (α + β) = m
Substitute (#4) into this ,
α + β + (12 - αβ) / (α + β) = 3n
4α + 4β + 4(12 - αβ) / (α + β) = 12n ---(#6)

Substitute (#5) into (#3) ,
αβ(12 - αβ) / (α + β) = 4n
3αβ(12 - αβ) / (α + β) = 12n ---(#7)

From (#6) and (#7) ,
4α + 4β + 4(12 - αβ) / (α + β) = 3αβ(12 - αβ) / (α + β)
(4α + 4β)(α + β) + 4(12 - αβ) = 3αβ(12 - αβ)
4α^2 + 8αβ + 4β^2 + 48 - 4αβ = 36αβ - 3α^2β^2
(3α^2 + 4)β^2 - 32αβ + 4α^2 + 48 = 0
β = (1/2)(1/(3α^2 + 4))[32α ± √(1024α^2 - 4(3α^2 + 4)(4α^2 + 48)) ] ---(#8)

D = 1024α^2 - 4(3α^2 + 4)(4α^2 + 48)
.. = 1024α^2 - 4(12α^4 + 160α^2 + 192)
.. = -48α^4 + 384α^2 - 768
.. = -48(α^4 - 8α^2 + 16)
.. = -48(α^2 - 4)^2
So , D becomes non-negative (zero) only if α = 2 . (α is positive .)
Substitute α = 2 into (#8) ,
β = (1/2)(1/(12 + 4))[64 ± √0 ]
.. = 2
Substitute α,β = 2 into (#2) ,
4 + 2γ + 2γ = 12
γ = 2
From (#1) and (#3) ,
m = α + β + γ = 6
n = αβγ / 4 = 2
Therefore m + n = 8 . (I think , your answer 2 is a value of α,β,γ .)

Too long