With usual notation in a - ABC 1-r1-1-r2 1-r2-1-r3 1-r3-1-r1 -K R3-a2b2c2 then K has value equal to-

The correct option is D 64 Given nbsp; Δ ABC ( 1/r_1+1/r_2 ) ( 1/r_2+1/r_3 ) ( 1/r_3+1/r_1 )=K R^3/a^2b^2c^2 nbsp;Using r_ 1 =Δ nbsp;( s a ...

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With usual no...

Question

With usual notation in a $Δ A B C (\frac{1}{r_{1}} + \frac{1}{r_{2}}) (\frac{1}{r_{2}} + \frac{1}{r_{3}}) (\frac{1}{r_{3}} + \frac{1}{r_{1}}) = \frac{K R^{3}}{a^{2} b^{2} c^{2}}$
then $K$ has value equal to?

1

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64

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128

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△ A B C,

(\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}},

K

(\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{64 R^{3}}{a^{2} b^{2} c^{2}}

1 + r + r^{2} + r^{3} + . . . . . . . + r^{n} = (1 + r) (1 + r^{2}) (1 + r^{4}) (1 + r^{8})

r_{1}, r_{2}, r_{3}

\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}

△ A B C

(1 - \frac{r_{1}}{r_{2}}) (1 - \frac{r_{1}}{r_{3}}) = 2

△ A B C

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