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mathmath333

  • one year ago

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  1. Loser66
    • one year ago
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    Typo is there, right? x =332, y = 333, not x =333, right?

  2. mathmath333
    • one year ago
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    good catch

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \text{If}\ x=332,\ y=333,\ z=335 \hspace{.33em}\\~\\ & \text{then}\ (x^{3}+y^{3}+z^{3}-3xyz) \ \text{is equal to}\hspace{.33em}\\~\\~\\ & a.)\ 9000 \hspace{.33em}\\~\\ & b.)\ 10000 \hspace{.33em}\\~\\ & c.)\ 7000 \hspace{.33em}\\~\\ & d.)\ 8000 \hspace{.33em}\\~\\ \end{align}}\) I have to solve this within a minute

  4. Loser66
    • one year ago
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    Let's wait for @imqwerty

  5. imqwerty
    • one year ago
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    a^3 + b^3 + c^3 – 3abc = (a+b+c)((a-b)^2 + (b-c)^2 + (c-a)^2) /2 use this

  6. mathmath333
    • one year ago
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    wow how did he get that thing

  7. mathmath333
    • one year ago
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    the traditional method may take an hour

  8. mathmath333
    • one year ago
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    or their might be some other trick using \(y=x+1,z=x+2\)

  9. Loser66
    • one year ago
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    What is the traditional method?

  10. mathmath333
    • one year ago
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    (traditional method)putting all those values and calculating

  11. imqwerty
    • one year ago
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    a^3 + b^3 +c^3 = (a+b+c)(a^2 +b^2 +c^2 -ab -bc -ca) =(a+b+c)(a^2 +a^2 +b^2 +b^2 +c^2 +c^2 -2ab-2bc-2ca)/2 did u get it now

  12. mathmath333
    • one year ago
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    yes i always forget the formula within 2 days

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