If the equation whose roots are the squares of the roots of the cubic x3−ax2+bx−1=0  is identical with the given cubic equation, then

Given equation isx3−ax2+bx−1=0. If roots of the equation be α,β,γ,  then α2+ β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)                     =a2−2b α2β2+γ2β2+γ2α2=(αβ+βγ+γα)2−2αβγ(α+β+γ)                                  =b2−2a                     α2β2γ2=1 So the equation whose roots are  is given by x3−(a2−2b)x2+(b2−2a)x−1=0 It is identical to  x3−ax2+bx−1=0 ⇒  a2−2b=a  and b2−2a=b Eliminating b, we get (a2−a)24−2a=a2−a2 ⇒  a{a(a−1)2−8−2(a−1)}=0 ⇒ a(a3−2a2−a−6)=0 ⇒  a(a−3)(a2+a+2)=0 ⇒a=0  or a=3  or a2+a+2=0, which gives b=0  or b=3  orb2+b+2=0.  So, a=b=0  or a=b=3  ora, b  are roots of x2+x+2=0.