Question

p(q-r)+q(r-p)+r(p-q)=0
Should prove that LHS=RHS
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Answer 1

Step-by-step explanation:

$LHS\\\\ =p(q-r)+q(r-p)+r(p-q) \\\\ = pq-pr+qr-pq+pr-qr \\\\ =0 \\\\ = RHS$

Answer 2

Answer:

$lhs \\ \\ p(q - r) + q(r - p) + r(p - q) \\ \\ = pq - pr + qr - pq + pr - qr \: \\ \\ so \: everthying \: gets \: cancelled \\ \\ = 0 \\ \\ and \: rhs \: = 0$

hope it helps u ✌