Simplify :
(a+b+c)(a−1b−1+b−1c−1+c−1a−1)\dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})}(a−1b−1+b−1c−1+c−1a−1)(a+b+c)
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Given,
Simplifying the expression :
⇒(a+b+c)(a−1b−1+b−1c−1+c−1a−1)⇒(a+b+c)(1ab+1bc+1ac)⇒(a+b+c)(c+a+babc)⇒((a+b+c)×abca+b+c)⇒abc.\Rightarrow \dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} \\[1em] \Rightarrow \dfrac{(a + b +c)}{\Big(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac}\Big)} \\[1em] \Rightarrow \dfrac{(a + b + c)}{\Big(\dfrac{c + a + b}{abc}\Big)} \\[1em] \Rightarrow {\Big(\dfrac{(a + b + c) \times abc}{a + b + c}\Big)} \\[1em] \Rightarrow abc.⇒(a−1b−1+b−1c−1+c−1a−1)(a+b+c)⇒(ab1+bc1+ac1)(a+b+c)⇒(abcc+a+b)(a+b+c)⇒(a+b+c(a+b+c)×abc)⇒abc.
Hence, (a+b+c)(a−1b−1+b−1c−1+c−1a−1)=abc\dfrac{(a + b +c)}{(a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1})} = abc(a−1b−1+b−1c−1+c−1a−1)(a+b+c)=abc.
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(a-1 × b-1) ÷ (a-1 + b-1)
(a + b)-1 × (a-1 + b-1)
Prove that:
(xaxb)a+b×(xbxc)b+c×(xcxa)c+a=1\Big(\dfrac{x^a}{x^b}\Big)^{a + b} \times \Big(\dfrac{x^b}{x^c}\Big)^{b + c} \times \Big(\dfrac{x^c}{x^a}\Big)^{c + a} = 1(xbxa)a+b×(xcxb)b+c×(xaxc)c+a=1
(xaxb)1ab×(xbxc)1bc×(xcxa)1ac=1\Big(\dfrac{x^a}{x^b}\Big)^{\dfrac{1}{ab}} \times \Big(\dfrac{x^b}{x^c}\Big)^{\dfrac{1}{bc}} \times \Big(\dfrac{x^c}{x^a}\Big)^{\dfrac{1}{ac}} = 1(xbxa)ab1×(xcxb)bc1×(xaxc)ac1=1
