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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

a,b,c are positive integers

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

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- Thread starter
- #1

- Jan 25, 2013

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a,b,c are positive integers

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

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- Feb 14, 2012

- 3,963

My first solution:a,b,c are positive integers

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!

Second solution:

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.

Last edited:

- Mar 31, 2013

- 1,358

Asked by you July 19 2013 and answered by mea,b,c are positive integers

(a-b)(b-c)(c+a)=- 90

(a-b)(b+c)(c-a)=42

(a+b)(b-c)(c-a)=- 60

find a,b,c

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- #4

Here is the link:Asked by you July 19 2013 and answered by me

http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-5637.html

Given all the problems generously posted by some of our members, this is bound to happen from time to time.

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- #5

- Feb 14, 2012

- 3,963

Asked by you July 19 2013 and answered by me

Oh My...I was totally unaware of that challenge problem been posted before at our site, and even if I did see it before, I didn't remember it at all.Here is the link:

http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-5637.html

Given all the problem generously posted by some of our members, this is bound to happen from time to time.

Fortunately my solutions don't match with either

- Mar 31, 2013

- 1,358

My first solution:

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!

Second solution:

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.

we know we must have c>a>b>0

I do not know the basis for the same.

I do not know the basis for the same.