In triangle , if1r1+1r21r2+1r31r3+1r1=KR3a2b2c2 then k=

1r1+1r21r2+1r31r3+1r1=s−aΔ+s−bΔs−bΔ+s−cΔs−cΔ+s−aΔ=2s−a−bΔ2s−b−cΔ2s−c−aΔ=abcΔ3=4RΔΔ3=4RΔ24=4Rabc4R2=64R3a2b2c2

In triangle , if1r1+1r21r2+1r31r3+1r1=KR3a2b2c2 then k=

1r1+1r21r2+1r31r3+1r1=s−aΔ+s−bΔs−bΔ+s−cΔs−cΔ+s−aΔ=2s−a−bΔ2s−b−cΔ2s−c−aΔ=abcΔ3=4RΔΔ3=4RΔ24=4Rabc4R2=64R3a2b2c2

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In triangle , if
$(\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}}$ then k=

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answer is C.

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Detailed Solution

$\begin{array}{l} (\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) \\ = (\frac{s - a}{Δ} + \frac{s - b}{Δ}) (\frac{s - b}{Δ} + \frac{s - c}{Δ}) (\frac{s - c}{Δ} + \frac{s - a}{Δ}) \\ = (\frac{2 s - a - b}{Δ}) (\frac{2 s - b - c}{Δ}) (\frac{2 s - c - a}{Δ}) \\ = \frac{a b c}{Δ^{3}} = \frac{4 R Δ}{Δ^{3}} = \frac{4 R}{Δ^{2}} 4 = \frac{4 R}{{(\frac{a b c}{4 R})}^{2}} = \frac{64 R^{3}}{a^{2} b^{2} c^{2}} \end{array}$