This is a problem from the book -“অলিম্পিয়াড সমগ্র”।
তিনটি সংখ্যার যোগফল 6, তাদের বগসমূহের যোগফল 8 এবং তাদের কিউবের যোগফল 5। তাদের চতুথ পাওয়ারের যোগফল কত?
Fourth Power
- Fahim Shahriar
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Name: Fahim Shahriar Shakkhor
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Re: Fourth Power
Quite interesting one.
Solution:
Solution:
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Phlembac Adib Hasan
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Re: Fourth Power
Nice solution.It can be done also using another approach (Not as beautiful).As we know $a+b+c$, $ab+bc+ca$ and $abc$, we can make a cubic equation $x^3-6x^2+14x-\frac {41}{3}=0$ by solving it the problem (brutally) gets killed.
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- Fahim Shahriar
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- Joined:Sun Dec 18, 2011 12:53 pm
Re: Fourth Power
Very nice solution and easy too. @Sourav Das
While solving it by cubic equation, we get-
$x^3-6x^2+14x-\frac{41}{3}=0$
Multiplying it by x,
$x^4-6x^3+14x^2-\frac{41x}{3}=0$
Now,
$a^4-6a^3+14a^2-\frac{41a}{3}=0$
$b^4-6b^3+14b^2-\frac{41b}{3}=0$
$c^4-6c^3+14c^2-\frac{41c}{3}=0$
Adding them, we have>
$(a^4+b^4+c^4)-6(a^3+b^3+c^3)+14(a^2+b^2+c^2)-41(a+b+c)/3=0$
Therefore, $(a^4+b^4+c^4)=6*5+14*8-41*6/3=0$
While solving it by cubic equation, we get-
$x^3-6x^2+14x-\frac{41}{3}=0$
Multiplying it by x,
$x^4-6x^3+14x^2-\frac{41x}{3}=0$
Now,
$a^4-6a^3+14a^2-\frac{41a}{3}=0$
$b^4-6b^3+14b^2-\frac{41b}{3}=0$
$c^4-6c^3+14c^2-\frac{41c}{3}=0$
Adding them, we have>
$(a^4+b^4+c^4)-6(a^3+b^3+c^3)+14(a^2+b^2+c^2)-41(a+b+c)/3=0$
Therefore, $(a^4+b^4+c^4)=6*5+14*8-41*6/3=0$
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College