Show that: $$(p-q)(p+q)+ (q-r)(q+r)+(r-p)(r+p) = 0$$
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Solution
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Taking LHS, we have LHS = $$(p-q)(p+q)+ (q-r)(q+r)+(r-p)(r+p) $$ $$ = p^2 - q^2 + q^2 - r^2 + r^2 - p^2 \ \ \ \ \ [ Using, \ (a + b) (a - b) = a^2 - b^2]$$ $$ = 0 = RHS$$
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