333nt sure bt itz btwn 8 and 27...
\hspace{-16}$If $\bf{a,b}$ and $\bf{c}$ are different real no. such that\\\\\\ \begin{Vmatrix} \bf{a^3=3b^2+3c^2-25} \\\\ \bf{b^3=3a^2+3c^2-25} \\\\ \bf{c^3=3a^2+3b^2-25} \end{Vmatrix}\\\\\\ Then find value of $\bf{abc=}$
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3 Answers
2305Without loss of generality we may assume c≤a≤b.
Let c = a-n & b = a+m where m,n≥0.
Using elementary algebra we can find that,
3b2+3c2-a3 ≥ 3a2+3c2-b3
This is because-
3(a+m)2 + 3(a-n)2 - a3 ≥ 3a2+3(a-n)2-(a+m)3
where equality holds only when m=n=0.
Since in the above equations the RHS and the LHS of the inequality are both equal to 25.
m=n=0 or a=b=c
Therefore,
3b2+3c2-a3
=3a2+3a2-a3 =25
or, a=5
Therfore a=b=c=5
- Hardik Sheth yeah...gr8Upvote·0· Reply ·2013-03-12 02:26:57
1057@shashwata...see the first line of the question....a=b=c=5 is obviously not the solution man111 wanted...i tried wolfram alpha....it's showing abc=2....
