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shubhamsrg

  • 3 years ago

if (b^4) + (1/b)^4 =6 then find the value of (b+ i/b)^16 where i = sqrt(-1)

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  1. shubhamsrg
    • 3 years ago
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    @mukushla ??

  2. mukushla
    • 3 years ago
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    use this \[(b^2-\frac{1}{b^2})^2+2=b^4+\frac{1}{b^4}=6\]\[(b^2-\frac{1}{b^2})^2=4\]\[b^2-\frac{1}{b^2}=(b+\frac{i}{b})^2-2i=\pm2\]

  3. shubhamsrg
    • 3 years ago
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    omg!! buddy,,which planet are you from? seriously??

  4. shubhamsrg
    • 3 years ago
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    thanks a lot!

  5. mukushla
    • 3 years ago
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    lol....yw :)

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