Multiple Choice Question (MCQ)

The greatest integer less than or equal to (1+\sqrt{2} )^6

  1. ×

    196

  2. 197

  3. ×

    198

  4. ×

    199

Collected in the board: The сube Problems

Steven Zheng posted 1 day ago

Answer

  1. (1+\sqrt{2} )^6

    =[(1+\sqrt{2} )^2]^3

    =(3+2\sqrt{2} )^3

    Let

    a=3+2\sqrt{2}
    (1)
    b=3-2\sqrt{2}
    (2)

    Then

    a+b=6
    (3)
    ab = 1
    (4)

    According to Vieta's formulas, a,b are two real roots of the quadratic equation

    x^2-6x+1 = 0

    Then

    a^2 = 6a-1

    b^2 = 6b-1

    a^2+b^2 = 6(a+b)-2 = 34
    (5)

    Using sum of cubes identity and substituting (3), (4), (5) gives

    a^3+b^3 = (a+b)(a^2-ab+b^2)

    =6\times(34-1)

    =198

    Therefore,

    a^3 = 198 - b^3

    Since 0< b< 1, 0< b^3< 1

    The greatest integer less than a^3 is 197

    B is the correct answer.

Steven Zheng posted 1 day ago

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