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x^2 +/- 1/x^2 given: find x^3 +/- 1/x^3

This type of question, asked 3 times: 21 Jul 2023 S3, 21 Jul 2023 S4, 17 Sep 2024 S1. All 3 questions below, oldest first.

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  1. Question 1

    21 Jul 2023 S3

    74
    If (x2−1x2)=46\left(x^2 - \frac{1}{x^2}\right) = 4\sqrt{6}, and x>1x > 1, what is the value of (x3−1x3)\left(x^3 - \frac{1}{x^3}\right)?

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  2. Question 2

    21 Jul 2023 S4

    74
    If (x2+1x2)=18(x^2 + \frac{1}{x^2}) = 18, and x>0x > 0, what is the value of (x3+1x3)(x^3 + \frac{1}{x^3})?

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  3. Question 3

    17 Sep 2024 S1

    56
    If x>1x > 1 and x2+1x2=83x^2 + \frac{1}{x^2} = 83, then x3−1x3x^3 - \frac{1}{x^3} is:

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