If α,β,γand δare the solution of the equation tanθ+π4=3 tan3θ, no two of which have equal tangents then the value of tanα+tanβ+tanγ+tanδ=

If α,β,γ and δare the solutions of the equation tanθ+π4=3tan3θ⇒1+tanθ1−tanθ=3×3tanθ−tan3θ1−3 tan2θ⇒1+t1−t=33t−t31−3t2    (putting t=tanθ)⇒3t4−6t2+8t−1=0∴sum of roots = t1+t2+t3+t4=0⇒tanα+tanβ+tanγ+tanδ=0