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Mathematics

If x is a positive real number and exponents are rational numbers, then simplify the following :

(xbxc)b+c−a(xcxa)c+a−b(xaxb)a+b−c\Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c}

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Answer

Given,

⇒(xbxc)b+c−a(xcxa)c+a−b(xaxb)a+b−c⇒(xb−c)b+c−a.(xc−a)c+a−b.(xa−b)a+b−c⇒x(b−c)(b+c−a).(x)(c−a)(c+a−b).(x)(a−b)(a+b−c)⇒xb2+bc−ba−cb−c2+ca.(x)c2+ca−cb−ac−a2+ab.(x)a2+ab−ac−ba−b2+bc⇒xb2−b2+bc−cb−cb+bc−ba+ab+ab−ba−c2+c2+ca+ca−ac−ac−a2+a2⇒x0=1.\Rightarrow \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \\[1em] \Rightarrow (x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b}.(x^{a - b})^{a + b - c} \\[1em] \Rightarrow x^{(b - c)(b + c - a)}.(x)^{(c - a)(c + a - b)}.(x)^{(a - b)(a + b - c)} \\[1em] \Rightarrow x^{b^2 + bc - ba - cb - c^2 + ca}.(x)^{c^2 + ca - cb - ac - a^2 + ab}.(x)^{a^2 + ab - ac - ba - b^2 + bc} \\[1em] \Rightarrow x^{b^2 - b^2 + bc - cb - cb + bc - ba + ab + ab - ba - c^2 + c^2 + ca + ca - ac - ac - a^2 + a^2} \\[1em] \Rightarrow x^0 = 1.

Hence, (xbxc)b+c−a(xcxa)c+a−b(xaxb)a+b−c\Big(\dfrac{x^b}{x^c}\Big)^{b + c - a}\Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} = 1.

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