If every pair from among the equations x^2 + px + qr = 0, x^2 + qx + rp and x^2 + rx + pq has only one common root, prove that the product of the three common roots is pqr.

Lets assume the common roots first.

Let -

x^2 + px + qr = 0 and x^2 + qx + rp = 0 have common root j;
x^2 + qx + rp = 0 and x^2 + rx + pq = 0 have common root k;
x^2 + rx + pq = 0 and x^2 + px + qr = 0 have common root l.
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From the first assumption,

j^2 + pj + qr = 0 and j^2 + qj + rp = 0

subtract these two equations,

(j^2 + pj + qr) - (j^2 + qj + rp) = 0 - 0

=> pj + qr - qj - rp = 0

You can see that two terms have j and two terms have r in common

=> pj - qj + qr - rp = 0
=> j(p - q) - r(p - q) = 0
=> (j - r)(p - q) = 0
=> j - r = 0; p - q = 0
=> j = r, p = q

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The second assumption
k^2 + qk + rp = 0  and k^2 + rk + pq = 0

Subtracting these two

=> k(q - r) - p(q - r) = 0
=> (k - p)(q - r) = 0
=> k = p, q = r
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The Third assumption
 l^2 + rl + pq = 0 and l^2 + pl + qr = 0

Subtracting these two  

=> l(r - p) - q(r - p) = 0
=> (l - q)(r - p) = 0
=> l = q, r = p
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You can see that if p = q, q = r and r = p becomes true, then p = q = r. Then, all the three equations will become the same. Then obviously, they'll all have same roots, and the concept of each pair having only one common root will fail. So results p = q, q = r and r = p will be discarded. We'll take the result j = r, k = p and l = q.

NOTE* - We could have cancelled (p - q), (q - r) and (r - p) from the equations formed after doing the subtractions, because we could've been able to foretell that they'll come as p = q = r in the future which can't not true, that means (p - q), (q - r) and (r - p) are all not equal to 0, and a term which is not equal to 0 can be cancelled by dividing the equation's both sides by it.

We've had assumed the common roots to be j, k, and l. We now know that j = r, k =p, and l = q, so the product of the common roots i.e. jkl will be equal to pqr. Hence Proved.