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Mathematics

Simplify the following:

(xa+bxc)a−b.(xb+cxa)b−c.(xc+axb)c−a\Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a}

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Answer

Given,

⇒(xa+bxc)a−b.(xb+cxa)b−c.(xc+axb)c−a=(xa+b−c)a−b.(xb+c−a)b−c.(xc+a−b)c−a=x(a2−ab+ba−b2−ca+cb).x(b2−bc+cb−c2−ab+ac).xc2−ca+ac−a2−bc+ba=xa2−ab+ba−b2−ca+cb+b2−bc+cb−c2−ab+ac+c2−ca+ac−a2−bc+ba=xa2−a2−ab+ba−b2+b2−ca+ac+cb−bc+cb−bc−ab+ba−ca+ac−c2+c2=x0=1.\Rightarrow \Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a} \\[1em] = (x^{a + b - c})^{a - b}.(x^{b + c - a})^{b - c}.(x^{c + a - b})^{c - a} \\[1em] = x^{(a^2 - ab + ba - b^2 - ca + cb)}.x^{(b^2 - bc + cb - c^2 - ab + ac)}.x^{c^2 - ca + ac - a^2 - bc + ba} \\[1em] = x^{a^2 - ab + ba - b^2 - ca + cb + b^2 - bc + cb - c^2 - ab + ac + c^2 - ca + ac - a^2 - bc + ba} \\[1em] = x^{a^2 - a^2 - ab + ba - b^2 + b^2 - ca + ac + cb - bc + cb - bc - ab + ba - ca + ac - c^2 + c^2} \\[1em] = x^0 = 1.

Hence, (xa+bxc)a−b.(xb+cxa)b−c.(xc+axb)c−a\Big(\dfrac{x^{a + b}}{x^c}\Big)^{a - b}.\Big(\dfrac{x^{b + c}}{x^a}\Big)^{b - c}.\Big(\dfrac{x^{c + a}}{x^b}\Big)^{c - a} = 1.

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