Fourth Power

[Fahim Shahriar]

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?

[sourav das]

Quite interesting one.
Solution:

$ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=14$
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
Implies, $abc=\frac{41}{3}$

Now comes the bingo:
$(a^4+b^4+c^4)-(a+b+c)(a^3+b^3+c^3)+(ab+bc+ca)(a^2+b^2+c^2)$
$-abc(a+b+c)=0$
And which implies that: $a^4+b^4+c^4=0$

Please correct me if I'm wrong in calculation.

To know more about the bingo, google Newton’s identities.

[Phlembac Adib Hasan]

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.

[Fahim Shahriar]

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, $(a^4+b^4+c^4)=6*5+14*8-41*6/3=0$