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If a+2/b = b+2/c = c+2/a then find value of abc.
Solution for both conditions, a = b = c and other condition also.
Here one direct solution can be observed when a = b = c. All equations are the same so that it can be possible.
Let a = b = c = n.
Let a = b = c = n.
Then abc = n³.
But I am looking for another possible case where all terms are not the same.
There is a possible solution for this case as below.
Let,
\[a+\frac{2}{b}----(1)\]
\[b+\frac{2}{c}----(2)\]
\[c+\frac{2}{a}----(3)\]
From (1) and (2),
\[a+\frac{2}{b} = b+\frac{2}{c}\]
\[\Rightarrow (a-b)= \frac{2}{c}-\frac{2}{b}\]
\[\Rightarrow (a-b)= \frac{2(b-c)}{bc}\]
Similarly,
\[(b-c)= \frac{2(c-a)}{ca}\]
and,
\[(c-a)= \frac{2(a-b)}{ab}\]
Multiply the above three equations with their respective sides (LHS with LHS and RHS with RHS).
We get,
\[(a-b)(b-c)(c-a)= \frac{8(b-c)(c-a)(a-b)}{(abc)^2}\]
By cancelling common terms from both sides,
\[1 = \frac{8}{(abc)^2}\]
\[\Rightarrow (abc)^2=8\]
\[\Rightarrow abc=\pm2\sqrt{2}\]
By this we can conclude:
If a = b = c = n,
abc = n³,
Else abc = ±2√2.
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