It is given that roots of the equation x3−6x2+11x−6=0 are α,β,γ Now, to find the equation whose roots are α2,β2,γ2, put α2=x⇒α=√x Since, α is the root of the given equation, so x3∕2−6x+11x1∕2−6=0 ⇒x1∕2(x+11)=6(x+1) On squaring both sides, we get x(x+11)2=36(x+1)2 ⇒x[x2+22x+121]=36[x2+2x+1] ⇒x3+22x2+121x=36x2+72x+36 ⇒x3−14x2+49x−36=0