Question

The sum of the squares of three negative numbers p,q,r equals the sum of the products of all possible pairs of the three numbers. Which of the following equals p3qr2 ?

1) (pq + qr + rp)3
2) q2(p3r + rp3)/2

Answer 1

@ganeshie8

Answer 2

we're given :
\(p^2+q^2+r^2 = pq=qr=rp\)

so, \(p^3qr^2 = pq*(pr)^2 = (p^2+q^2+r^2)^3\)

Answer 3

Right answer id option 2 @ganeshie8

Answer 4

"The sum of the squares of three negative numbers p,q,r equals the *sum* of the products of all possible pairs of the three numbers."
I think we are given that\[p^2+q^2+r^2=pq+qr+rp\]I could be wrong; I'm a bit rusty...

Answer 5

yes, ^^

Answer 6

Which of the following equals p3qr2 ?@TuringTest

Answer 7

I'm pretty bad at these problems, but\[p^2=pq+qr+rp-q^2-r^2\]\[r^2=pq+qr+rp-q^2-p^2\]\[pq=p^2+q^2+r^2-qr-rp\]multiplying all that out should give an answer, but I'm sure there's a smarter way to do it.

Answer 8

Thank you :)

Answer 9

You're welcome

Answer 10

something is off..

the answer is q2(p3r + rp3)/2 <<p^3 r and r p^3 are same!!>>


which means
p^3 q r^2 = q^2(p^3 r + rp^3)/2 holds true

p^3 and q can be factored out from the right
and gets cancelled,

r^2 = q (r+r)/2 = rq

r= q holds true
...