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Mathematics

Prove that :

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

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Answer

Solving L.H.S. of the given equation :

⇒(xaxb)a+b−c(xbxc)b+c−a(xcxa)c+a−b=(xa−b)a+b−c.(xb−c)b+c−a.(xc−a)c+a−b=x(a−b)(a+b−c).x(b−c)(b+c−a).x(c−a)(c+a−b)=xa2+ab−ac−ab−b2+bc.xb2+bc−ab−bc−c2+ac.xc2+ac−bc−ac−a2+ab=xa2−b2−ac+bc.xb2−c2−ab+ac.xc2−a2−bc+ab=xa2−b2−ac+bc+b2−c2−ab+ac+c2−a2−bc+ab=xa2−a2−b2+b2−c2+c2−ac+ac+bc−bc−ab+ab=x0=1.\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} \\[1em] = (x^{a - b})^{a + b - c}.(x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b} \\[1em] = x^{(a - b)(a + b - c)}.x^{(b - c)(b + c - a)}.x^{(c - a)(c + a - b)} \\[1em] = x^{a^2 + ab - ac - ab - b^2 + bc}.x^{b^2 + bc - ab - bc - c^2 + ac}.x^{c^2 + ac - bc - ac - a^2 + ab} \\[1em] = x^{a^2 - b^2 - ac + bc}.x^{b^2 - c^2 - ab + ac}.x^{c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - b^2 - ac + bc + b^2 - c^2 - ab + ac + c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - a^2 - b^2 + b^2 - c^2 + c^2 - ac + ac + bc - bc - ab + ab} \\[1em] = x^0 \\[1em] = 1.

Since, L.H.S. = R.H.S. = 1.

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

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