Let α, β, γ are the roots of the cubic equation a0 x3+3 a1 x2+3 a2 x+a3=0 (a0 ≠ 0). Then the value of (α-β)2+(β-γ)2+(γ-α)2 equals

Question

Let α, β, &gamma; are the roots of the cubic equation a0 x3+3 a1 x2+3 a2 x+a3=0 (a0 ≠ 0). Then the value of (α-β)2+(β-&gamma;)2+(&gamma;-α)2 equals (A) (18( a 22- a 0 a 1)/ a 02) (B) (18( a 22+ a 0 a 1)/ a 02) (C) (18( a 02- a 1 a 2)/ a 02) (D) (18( a 12- a 0 a 2)/ a 02)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/5f6f1110d22433e98509a56836143f4d-.png" /> Sigma α=α+β+&gamma;=(-3 a1/a0) Sigma α β=α β+β &gamma;+&gamma; α=(3 a2/a0) text Now, (α-β)2+(β-&gamma;)2+(&gamma;-α)2=2(α2+β2+&gamma;2)-2 Sigma α β=2((α+β+&gamma;)2-2 Sigma α β)-2 Sigma α β =2(α+β+&gamma;)2-6 Sigma α β=2((-3 a1/a0))2-6((3 a2/a0))=(18 a12/a02)-(18 a2/a0)=(18(a12-a0 a2)/a02)

Q. Let $α, β, γ$ are the roots of the cubic equation $a_{0} x^{3} + 3 a_{1} x^{2} + 3 a_{2} x + a_{3} = 0 (a_{0} \neq = 0)$ . Then the value of $(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2}$ equals

$\frac{18 ( a _{2}^{2} - a _{0} a _{1} )}{a _{0}^{2}}$

$\frac{18 ( a _{2}^{2} + a _{0} a _{1} )}{a _{0}^{2}}$

$\frac{18 ( a _{0}^{2} - a _{1} a _{2} )}{a _{0}^{2}}$

$\frac{18 ( a _{1}^{2} - a _{0} a _{2} )}{a _{0}^{2}}$

Q. Let α,β,γ are the roots of the cubic equation a0​x3+3a1​x2+3a2​x+a3​=0(a0​=0). Then the value of (α−β)2+(β−γ)2+(γ−α)2 equals

a02​18(a22​−a0​a1​)​

a02​18(a22​+a0​a1​)​

a02​18(a02​−a1​a2​)​

a02​18(a12​−a0​a2​)​

Q. Let $α, β, γ$ are the roots of the cubic equation $a_{0} x^{3} + 3 a_{1} x^{2} + 3 a_{2} x + a_{3} = 0 (a_{0} \neq = 0)$ . Then the value of $(α - β)^{2} + (β - γ)^{2} + (γ - α)^{2}$ equals

$\frac{18 ( a _{2}^{2} - a _{0} a _{1} )}{a _{0}^{2}}$

$\frac{18 ( a _{2}^{2} + a _{0} a _{1} )}{a _{0}^{2}}$

$\frac{18 ( a _{0}^{2} - a _{1} a _{2} )}{a _{0}^{2}}$

$\frac{18 ( a _{1}^{2} - a _{0} a _{2} )}{a _{0}^{2}}$