If a, b, c are distinct positive numbers, each different from 1, such that
$[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0, t h e n a b c =$

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Solution

The correct option is A 1
$[l o g_{b} a . l o g_{c} a - l o g_{a} a] + [l o g_{a} b . l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0 = [(l n a)^{3} + (I n b)^{3} + (I n c)^{3}] \frac{1}{I n a . I n b . I n c} - 3 = 0 = \frac{1}{I n a . l n b . I n c} [(I n a)^{3} + (I n b)^{3} + (I n c)^{3} - 3 I n a . I n b . I n c] = 0 = (I n a)^{3} + (I n b)^{3} + (I n c)^{3} - 3 I n a . I n b . I n c = 0 \Rightarrow I n a + I n b + I n c = 0 A s [a^{3} + b^{3} + c^{3} - 3 a b c = 0 \Rightarrow a + b + c = 0] \Rightarrow I n (a b c) = I n 1 ∴ a b c = 1.$

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[l o g_{b} a l o g_{c} a - l o g_{a} a] + [l o g_{a} b l o g_{c} b - l o g_{b} b] + [l o g_{a} c l o g_{b} c - l o g_{c} c] = 0

If $\frac{log a}{b - c} = \frac{log b}{c - a} = \frac{log c}{a - b}$ then prove that $a^{a} . b^{b} . c^{c} = 1$

[({log}_{b} a) ({log}_{c} b) ({log}_{a} c)]

log (\frac{a}{b - c}) = log (\frac{b}{c - a}) = log (\frac{c}{a - b})

a^{a} \times b^{b} \times c^{c} = 1

\frac{l o g a}{b - c} = \frac{l o g b}{c - a} = \frac{l o g c}{a - b}

a^{b + c} . b^{c + a} . c^{a + b}

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