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Question
Show that:
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
Solution
`(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)=1`
LHS = `(3^a/3^b)^(a+b)(3^b/3^c)^(b+c)(3^c/3^a)^(c+a)`
`=(3^(a-b))^(a+b)(3^(b-c))^(b+c)(3^(c-a))^(c+a)`
`=(3^((a-b)(a+b)))(3^((b-c)(b+c)))(3^((c-a)(c+a)))`
`=(3^(a^2-b^2))(3^(b^2-c^2))(3^(c^2-a^2))`
`=3^(a^2-b^2+b^2-c^2+c^2-a^2)`
`=3^0`
= 1
= RHS
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