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Mathematics

Show that: [xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

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Answer

To prove: [xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

Taking LHS:

[xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=[xa(a−b)x−b(a−b)].[xb(b−c)x−c(b−c)].[xc(c−a)x−a(c−a)]=[xa2−abx−ab+b2].[xb2−bcx−bc+c2].[xc2−acx−ac+a2]=[x(a2−ab)−(−ab+b2)].[x(b2−bc)−(−bc+c2)].[x(c2−ac)−(−ac+a2)]=[xa2−ab+ab−b2].[xb2−bc+bc−c2].[xc2−ac+ac−a2]=[xa2−b2].[xb2−c2].[xc2−a2]=x(a2−b2)+(b2−c2)+(c2−a2)=xa2−b2+b2−c2+c2−a2=x0=1=RHS\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a}\\[1em] = \Big[\dfrac{x^{a(a-b)}}{x^{-b(a-b)}}\Big].\Big[\dfrac{x^{b(b-c)}}{x^{-c(b-c)}}\Big].\Big[\dfrac{x^{c(c-a)}}{x^{-a(c-a)}}\Big]\\[1em] = \Big[\dfrac{x^{a^2-ab}}{x^{-ab+b^2}}\Big].\Big[\dfrac{x^{b^2-bc}}{x^{-bc+c^2}}\Big].\Big[\dfrac{x^{c^2-ac}}{x^{-ac+a^2}}\Big]\\[1em] = \Big[{x^{(a^2-ab)-(-ab+b^2)}}\Big].\Big[x^{(b^2-bc)-(-bc+c^2)}\Big].\Big[x^{(c^2-ac)-(-ac+a^2)}\Big]\\[1em] = \Big[{x^{a^2-ab+ab-b^2}}\Big].\Big[x^{b^2-bc+bc-c^2}\Big].\Big[x^{c^2-ac+ac-a^2}\Big]\\[1em] = \Big[x^{a^2-b^2}\Big].\Big[x^{b^2-c^2}\Big].\Big[x^{c^2-a^2}\Big]\\[1em] = x^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}\\[1em] = x^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] = x^0\\[1em] = 1 \\[1em] = \text{RHS}

∴ LHS = RHS

[xax−b]a−b.[xbx−c]b−c.[xcx−a]c−a=1\Big[\dfrac{x^a}{x^{-b}}\Big]^{a-b}.\Big[\dfrac{x^b}{x^{-c}}\Big]^{b-c}.\Big[\dfrac{x^c}{x^{-a}}\Big]^{c-a} = 1

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