Let f(x)=x3+ax2+bx+c=0
and α,β,γ be the roots of the above equation.
α+β+γ=−a
αβ+βγ+γα=b
αβγ=−c
f′(x)=3x2+2ax+b
Let x1,x2 be the roots of f′(x)=0
x1+x2=−2a3
x1x2=b3
Given, β is harmonic mean between x1 and x2.
⇒β=2x1x2x1+x2=2b−2a=b−a
⇒β=(αβ+βγ+γα)(α+β+γ)
⇒αβ+β2+βγ=αβ+βγ+γα
⇒β2=αγ
⇒α,β,γ are in G.P.
r=[2βα+γ]+[2αγαβ+βγ]
⇒r=[2√αγα+γ]+[2αγβ(α+γ)]
⇒r=[G.M.A.M.]+[H.M.G.M.]=0+0=0
{∵A.M.>G.M.>H.M.}
∴r∑i=1(i)3=3∑i=11=1+1+1=3