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Problem Of The Week #438 October 12th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,705
Here is this week's POTW:

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Determine all triples of real numbers $(a,\,b,\,c)$ such that

$abc=8\\ a^2b+b^2a+c^2a=73\\a(b-c)^2+b(c-a)^2+c(a-b)^2=98$

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,705
No one answered last week's POTW. (Sadface) However, you can read the suggested solution below:

Expanding the third equation we get

$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b-6abc=98$

Replace $a^2b+b^2a+c^2a$ by 73 and $abc=8$ into the above equation and rearrange and factorize, we get

$a^2c+b^2v+c^2b=a^2b+b^2a+c^2a\\a^2c+b^2v+c^2b-(a^2b+b^2a+c^2a)=0\\-(a-b)(b-c)(c-a)=0$
Therefore we know that at least two variables are the same.

WLOG, we let $a=c$, the first equation becomes $a^2b=8$ and the second equation becomes $a^2b+ab^2+a^3=73$ and third equation becomes $a(a-b)^2=49$.

Solving this system for $a$ results in

$8+\dfrac{64}{a^3}+a^3=73\\a^6-65a^3+64=0\\(a^3-64)(a^3-1)=0$
$\therefore a=c=1$ or $a=c=4$.

When $a=c=1$, $b=8$ or when $a=c=4$, $b=\dfrac{1}{2}$.

The solutions are hence $(a,\,b,\,c)=(1,\,8,\,1),\,\left(4,\,\dfrac{1}{2},\,4\right)$ and their permutations.
 
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