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If a = 1 + "log"(x) yz, b = 1 + "log"(y)...

If `a = 1 + "log"_(x) yz, b = 1 + "log"_(y) zx, c =1 + "log"_(z) xy,` then ab+bc+ca =

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We oberve that
` a = 1 log_(x) yz = log_(x)x + log_(x)yz= log_(x) xyz = (1)/(log_(xyz^(x)))`
`rArr (1)/(a) = log_(xyz^(x))`
Similarly , ` (1)/(b) = log_(xyz) y and (1)/(c) = log_(xyz^(z))`
Now , ` (1)/(a) + (1)/(b) + (1)/(c) = log_(xyz) x +log_(xyz) y + log_(xyz) z = log_(xyz) xyz = 1 `
`rArr ab + bc + ca = abc `
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