prove

prove the expression is non zero if p≠q≠r

pq(p-q) + qr(q-r) + rp(r-p)

sort of doubt actually :)

5 Answers

21
Shubhodip ·

my solution : if the given expression equals zero (p,p2) , (q,q2) , (r,r2) will be collinear

p,p2 ,q,q2 ,r,r2 are distinct points on the parabola y= x2

but a straight line intersects a parabola at maximum 2 points

so the given is nonzero

i am looking for a basic algebric proof

71
Vivek @ Born this Way ·

(p,p2) , (q,q2) , (r,r2)

How do you infere this please let me know.

341
Hari Shankar ·

Its true for the simple reason that \sum pq(p-q) = -(p-q)(q-r)(r-p)

Now, how did we arrive at the factorisation?

This is a well known technique. By setting p=q, we see that the expression equals zero

So, (p-q) must be a factor of LHS.

So (p-q)(q-r)(r-p) divides LHS. Comparing the degree, we see that

LHS = K (p-q)(q-r)(r-p) where K is a constant.

Comparing the coeff of p on both sides, K=-1 and we are done

21
Shubhodip ·

thank you very much sir thanks a lot

341
Hari Shankar ·

We can tie this with Shubhodip's remark about the points.

If the points are collinear then the are of the triangle formed by the three points is zero i.e.
\begin{vmatrix} 1 & p& p^2 \\ 1&q &q^2 \\ 1 & r & r^2 \end{vmatrix} = 0

Now look at the Vandermonde determinant

http://en.wikipedia.org/wiki/Vandermonde_matrix

You can now see why we must have p≠q≠r

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