Prove that, ((1/(a − b)) + (1/(b − c)) + (1/(c − a)))^2 = (1/((a − b)^2 )) + (1/((b − c)^2 )) + (1/((c − a)^2 ))


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Question Number 181722 by Agnibhoo98 last updated on 29/Nov/22

$Prove that,$ ${(\frac{1}{a - b} + \frac{1}{b - c} + \frac{1}{c - a})}^{2} =$ $\frac{1}{{(a - b)}^{2}} + \frac{1}{{(b - c)}^{2}} + \frac{1}{{(c - a)}^{2}}$
	Answered by Rasheed.Sindhi last updated on 29/Nov/22

	${(\frac{1}{a - b} + \frac{1}{b - c} + \frac{1}{c - a})}^{2} =$ $= \frac{1}{{(a - b)}^{2}} + \frac{1}{{(b - c)}^{2}} + \frac{1}{{(c - a)}^{2}}$ $a - b = p, b - c = q, c - a = r$ $p + q + r = 0$ $T o p r o v e :$ ${(\frac{1}{p} + \frac{1}{q} + \frac{1}{r})}^{2} = \frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}}$ $LHS :$ $\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} + 2 (\frac{1}{p q} + \frac{1}{q r} + \frac{1}{r p})$ $\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} + 2 (\frac{r}{p q r} + \frac{p}{p q r} + \frac{q}{p q r})$ $\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} + 2 (\frac{p + q + r}{p q r})$ $\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} + 2 (\frac{0}{p q r})$ $\frac{1}{p^{2}} + \frac{1}{q^{2}} + \frac{1}{r^{2}} = RHS$
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