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Mathematics

Prove that:

[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

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Answer

To prove:[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

Taking LHS:

[xaxb]1ab[xbxc]1bc[xcxa]1ca=[xa−b]1ab[xb−c]1bc[xc−a]1ca=[x]a−bab[x]b−cbc[x]c−aca=[x]aab−bab[x]bbc−cbc[x]cca−aca=[x]1b−1a[x]1c−1b[x]1a−1c=[x]1b−1a+1c−1b+1a−1c=[x]1b−1b−1a+1a+1c−1c=x0=1=RHS\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}}\\[1em] = \Big[x^{a-b}\Big]^{\dfrac{1}{ab}}\Big[x^{b-c}\Big]^{\dfrac{1}{bc}}\Big[x^{c-a}\Big]^{\dfrac{1}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{a-b}{ab}}\Big[x\Big]^{\dfrac{b-c}{bc}}\Big[x\Big]^{\dfrac{c-a}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{a}{ab}-\dfrac{b}{ab}}\Big[x\Big]^{\dfrac{b}{bc}-\dfrac{c}{bc}}\Big[x\Big]^{\dfrac{c}{ca}-\dfrac{a}{ca}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{a}}\Big[x\Big]^{\dfrac{1}{c}-\dfrac{1}{b}}\Big[x\Big]^{\dfrac{1}{a}-\dfrac{1}{c}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{c}}\\[1em] = \Big[x\Big]^{\dfrac{1}{b}-\dfrac{1}{b}-\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{c}-\dfrac{1}{c}}\\[1em] = x^0\\[1em] = 1 \\[1em] = \text{RHS}

∴ LHS = RHS

Hence,[xaxb]1ab[xbxc]1bc[xcxa]1ca=1\Big[\dfrac{x^a}{x^b}\Big]^{\dfrac{1}{ab}}\Big[\dfrac{x^b}{x^c}\Big]^{\dfrac{1}{bc}}\Big[\dfrac{x^c}{x^a}\Big]^{\dfrac{1}{ca}} = 1

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