Right, and in every proof above, at some point there is a possible
division by zero. Therefore the proof is not valid in cases where R or
P or Q are zero, and there are infinitely many such cases.
The problem states P+Q+R=0 as the only constraint. There are
infinitely many cases which fit that constraint where the expression
is not equal to 3.
Don

On Sep 15, 11:57 am, abhinav gupta <guptaabhinav...@gmail.com> wrote:
> u cnt divide a number by 0..that thing is self undrstod,,,,
>
> On Thu, Sep 15, 2011 at 9:49 AM, Piyush Grover 
> <piyush4u.iit...@gmail.com>wrote:
>
>
>
> > Don is right....
>
> > if R = 0, P = 1 and Q = -1 then the given expression is UNDEFINED!!!
>
> > On Thu, Sep 15, 2011 at 10:16 PM, abhinav gupta <guptaabhinav...@gmail.com
> > > wrote:
>
> >> Shut up...its 3,,
>
> >> On Thu, Sep 15, 2011 at 9:43 AM, Don <dondod...@gmail.com> wrote:
>
> >>> It might be 3, but it doesn't have to be 3.
> >>> Don
>
> >>> On Sep 14, 11:56 pm, NAGARAJAN SIVARAMAN <naga4...@gmail.com> wrote:
> >>> > if P+Q+R= 0  then P2 /QR  + Q2/PR + R2/PQ = ??
>
> >>> > how to solve this??
>
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