Show that p³q + q³r + r³p is a constant

[anemone]

Show that for all real numbers $p, q, r$ such that $p+q+r=0$ and $pq+pr+qr=-3$, the expression $p^3q+q^3r+r^3p$ is a constant.

 

[juantheron]

Re: Show that p³q+q³r+r³p is a constant

Here is my solution:

Spoiler

Given $p+q+r = 0$ and $pq+qr+rp = -3$ Now Let $pqr = k$

Now form an cubic equation whose roots are $x = p\;,q\;,r$

$x^3-(p+q+r)x^2+(pq+qr+rp)x-pqr = 0$

$x^3-3x-k=0\Rightarrow x^3 = 3x+k$

Now If $x = p$ is a root of given equation, Then $p^3 = 3p+k$

Similarly If $x = q$ is a root of given equation, Then $q^3 = 3q+k$

Similarly If $x = r$ is a root of given equation, Then $r^3 = 3r+k$

So $p^3q = (3p+k)q = 3pq+kq$

Similarly $q^3r = (3q+k)r = 3pr+kr$

Similarly $r^3p = (3r+k)p = 3rp+kp$

Now Add all These, we get $p^3q+q^3r+r^3p = 3(pq+qr+rp)+(p+q+r)k = -9+0 = -9$(Constant)

 

[kaliprasad]

Re: Show that p³q+q³r+r³p is a constant

Not elegant but different approach

Spoiler

P = - (q+r)

So p^3 q = - p^2 q^2 – p^2 q r
Similarly q^3 r = q^2 r^2 – q^2 pr
r^3 p = r^2 p^2 – r^2 pr
add all 3 to get
p^3 q + q^3 r + r^3 p = - (p^2q^2 + q^2 r^2 + r^2 p^2) – pqr(p+q + r)
= - (p^2q^2 + q^2 r^2 + r^2 p^2) .. (1)
Now as we have p^2 q^2 + q^2 r^2 + r^2 p^2 above we square
pq+pr+qr=−3 to get
p^2q^2 + p^2 r^2 + q^2 r ^2 + 2p^2qr + 2r^2qp + 2 q^pr = 9
or p^2q^2 + p^2 r^2 + q^2 r ^2 + 2pqr(p + r + q) = 9
or p^2q^2 + p^2 r^2 + q^2 r ^2 = 9 ... (2)
from (1) and (2) p^3 q + q^3 r + r^3 p = - 9