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 is x3−ax2+bx−1=0. If roots of the equation be α,β,γ, thenα2+β2+γ2=(α+β+γ)2−2(αβ+βγ+γα)=a2−2b α2β2+β2γ2+γ2α2=(αβ+βγ+γα)2−2αβγ(α+β+γ)=b2−2a α2β2γ2=1So, the equation whose roots are α2,β2,γ2 is given byx3−a2−2bx2+b2−2ax−1=0It is identical tox3−ax2+bx−1=0⇒ a2−2b=a and b2−2a=bEliminating b, we geta2−a24−2a=a2−a2or   aa(a−1)2−8−2(a−1)=0or   aa3−2a2−a−6=0or   a(a−3)a2+a+2=0⇒  a=0 or a=3 or a2+a+2=0which gives b = 0 or b = 3 or b2 + b + 2 = 0. So, a = b = 0 or a = b = 3 or a, b are roots of x2 + x + 2 = 0.