Theory of EquationsmediumFree

Theory of Equations: Roots Equation 111x 11x Equals

JEE Maths question with a full step-by-step solution.

Question

\alpha, \beta, \gamma

are roots of the equation

111x^3 - 11x - 1 = 0

, then

(\alpha\beta)^{-2} + (\beta\gamma)^{-2} + (\gamma\alpha)^{-2}

equals

2332

1331

1210

2442

correct

Solution

Step 1: By Vieta's:

\alpha+\beta+\gamma = 0

\alpha\beta+\beta\gamma+\gamma\alpha = -\dfrac{11}{111}

\alpha\beta\gamma = \dfrac{1}{111}

. Step 2:

(\alpha\beta)^{-2}+(\beta\gamma)^{-2}+(\gamma\alpha)^{-2} = \frac{\gamma^2+\alpha^2+\beta^2}{(\alpha\beta\gamma)^2}

Step 3:

\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 - 2\left(-\frac{11}{111}\right) = \frac{22}{111}

Step 4:

\frac{22/111}{(1/111)^2} = \frac{22}{111} \cdot 111^2 = 22 \times 111 = 2442

Correct answer: (4)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.