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If α and β be two roots of the equation x^2-64x+256=0. Then the value of ( α ^3 β ^5)^ 1 8+( β ^3 α ^5)^ 1 8 is :

JEE Main 2020 — Mathematics Algebra

2020mcqeasy

If α\alphaα and β\betaβ be two roots of the equation x2−64x+256=0.{x}^{2}-64x+256=0.x2−64x+256=0. Then the value of (α3β5)18+(β3α5)18{(\frac{{\alpha }^{3}}{{\beta }^{5}})}^{\frac{1}{8}}+{(\frac{{\beta }^{3}}{{\alpha }^{5}})}^{\frac{1}{8}}(β5α3​)81​+(α5β3​)81​ is :

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

Options

  1. A

    222

  2. B

    333

  3. C

    111

  4. D

    444

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