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hartnn

Anyone getting bored? try this : If a³ + b³ = 0 , then the value of log (a + b) – 0.5(log a + log b + log 3) is equal to ??

  • one year ago
  • one year ago

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  1. hartnn
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    just for safety: a\(\ne\)b a\(\ne\)-b

    • one year ago
  2. ParthKohli
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    \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]So,\[(a + b)(a^2 - ab + b^2) = 0\]Dividing both sides by a + b,\[a^2 - ab + b^ 2 = 0\]Since \(a^2 - 2ab + b^2 = (a - b)^2\), we have,\[a^2 - 2ab + b^2 +ab = 0\]\[(a - b)^2 + ab = 0\]\[(a - b)^2 = - ab\]\[a^2 + b^2 = ab\]One such case is a = 0 and b = 0, but you haven't allowed that. :(

    • one year ago
  3. hartnn
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    hmm....nice try no other formula for a^3+b^3 ?

    • one year ago
  4. nipunmalhotra93
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    Are you sure that your question is correct? If a^3 + b^3=0, then either a=-b or a=b=0.... :\ (because x^3 if a one one function)

    • one year ago
  5. hartnn
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    yes, question is correct.

    • one year ago
  6. nipunmalhotra93
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    0

    • one year ago
  7. hartnn
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    how ?

    • one year ago
  8. nipunmalhotra93
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    continuing from parthkohli, |dw:1347876944383:dw|

    • one year ago
  9. nipunmalhotra93
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    and your question simplifies to |dw:1347876996875:dw|

    • one year ago
  10. mukushla
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    neat solution

    • one year ago
  11. hartnn
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    yes, thats right, its 0.there is easier way. rather than continuing...

    • one year ago
  12. nipunmalhotra93
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    @mukushla thanks :) @hartnn what is it?

    • one year ago
  13. hartnn
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    \((a^3+b^3)=(a+b)^3-3ab(a+b)=0\) and the rest is ............

    • one year ago
  14. nipunmalhotra93
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    haha... yeah :D

    • one year ago
  15. nipunmalhotra93
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    well even though we solved it, it IS technically wrong I think? :\

    • one year ago
  16. nipunmalhotra93
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    |dw:1347877692079:dw|

    • one year ago
  17. nipunmalhotra93
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    |dw:1347877816341:dw|

    • one year ago
  18. hartnn
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    if i read it correctly, there exist no a,b not equal to 0, such that a^2+b^2-ab = 0.....??

    • one year ago
  19. nipunmalhotra93
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    Yea...to be clearer, I mean that a^2+b^2-ab=0 iff a=b=0

    • one year ago
  20. hartnn
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    u sure about a^2+b^2-ab=0 iff a=b=0 ?

    • one year ago
  21. nipunmalhotra93
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    yea... you can try that... you won't find any a, b s.t a,b are not both zero and the condition is satisfied....

    • one year ago
  22. nipunmalhotra93
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    algebraically, x^3, is a one one function... so, a^3 +b^3=0 implies a^3=(-b)^3 which implies a=-b

    • one year ago
  23. hartnn
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    a^3 =-b^3 have 3 roots.....i think

    • one year ago
  24. nipunmalhotra93
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    Graphically,

    • one year ago
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  25. nipunmalhotra93
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    I plotted two graphs.... z=xy and z=x^2+y^2 they intersect only at (0,0) Yes there are three values of b for every a. But they are complex.

    • one year ago
  26. nipunmalhotra93
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    If your question covers complex numbers, then it's fine.... but for real, this is it :)

    • one year ago
  27. nipunmalhotra93
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    @hartnn Thanks for the time man.. that was a good exercise for me ^_^

    • one year ago
  28. hartnn
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    hmm..log of complex numbers....didn't think of it b4 posting....and thanks to u for wonderful explanation.

    • one year ago
  29. nipunmalhotra93
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    you're most welcome buddy!

    • one year ago
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