1r - 1r1 1r - 1r2 1r - 1r3 - 16Rr2a - b - c2

To prove: ( 1 r 1 r _ 1 #160; ) ( 1 r 1 r _ 2 #160; ) ( 1 r 1 r _ 3 #160; ) = 16R r ^ 2 ( a+b+c ) #160; ^ 2 We know, r= ...

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(a + b)^2 Expansion and Visualisation

1r - 1r1 1r ...

Question

$(\frac{1}{r} - \frac{1}{r_{1}}) (\frac{1}{r} - \frac{1}{r_{2}}) (\frac{1}{r} - \frac{1}{r_{3}}) = \frac{16 R}{r^{2} (a + b + c)^{2}}$

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

1 + \frac{1}{r} + \frac{1}{r^{4}} + \frac{1}{r^{9}} + \frac{1}{r^{16}} + \dots = 1 + \frac{1}{r -} \frac{r}{r^{3} + 1 -} \frac{r^{3}}{r^{5} + 1 -} \frac{r^{5}}{r^{7} + 1 -} \dots

r

a [\frac{1}{r^{3}} + \frac{1}{r} + r + r^{3}] = 85

a = 8

S = {(x, y) \in R^{3} : \frac{y^{2}}{1 + r} - \frac{x^{2}}{1 - r} = 1}

r \neq \pm 1

S

1 / r_{1} + 1 / r_{2} + 1 / r_{3} = 1 / r

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