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Question

Show that:
$$(p-q)(p+q)+ (q-r)(q+r)+(r-p)(r+p) = 0$$

Solution
Verified by Toppr

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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