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- Feb 14, 2012

- 3,952

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,952

Here is my solution:

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)

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- Mar 31, 2013

- 1,356

Not elegant but different approach

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

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