Consider the given expression x3−6x2+11x−6=0So(x−1)(x−2)(x−3)‌‌=0 x‌‌=1,2,3 ‌ Therefore, ‌α=1,β‌‌=2,γ=3 Now, α2+β2‌‌=(12)+(22) ‌‌=5 ‌‌=α′(‌ say ‌)
β2+γ2‌‌=22+32 ‌‌=13 ‌‌=β′‌‌(‌ say ‌)
γ2+α2‌‌=32+1 ‌‌=10 ‌‌=γ′‌‌(‌ say ‌)
Thus, the expression of the equation that has roots α′,β′,γ′ is, x3−(α′+β′+γ′)+(α′β′+β′γ′+γ′α′)x−α′β′γ′=0 x3−(5+13+10)x2+(5×13+13×10+10×5)×−5×13×10=10 x3−28x2+245x−650=0