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Mathematics

Simplify the following:

(xax−b)a2−ab+b2.(xbx−c)b2−bc+c2.(xcx−a)c2−ca+a2\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2}

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Answer

Given,

⇒(xax−b)a2−ab+b2.(xbx−c)b2−bc+c2.(xcx−a)c2−ca+a2⇒(x(a−(−b)))(a2−ab+b2).(x(b−(−c)))(b2−bc+c2).(x(c−(−a)))(c2−ca+a2)⇒(x(a+b))(a2−ab+b2).(x(b+c))(b2−bc+c2).(x(c+a))(c2−ca+a2)\Rightarrow \Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} \\[1em] \Rightarrow (x^{(a - (-b))})^{(a^2 - ab + b^2)}.(x^{(b - (-c))})^{(b^2 - bc + c^2)}.(x^{(c - (-a))})^{(c^2 - ca + a^2)} \\[1em] \Rightarrow (x^{(a + b)})^{(a^2 - ab + b^2)}.(x^{(b + c)})^{(b^2 - bc + c^2)}.(x^{(c + a)})^{(c^2 - ca + a^2)}

By formula,

(x3 + y3) = (x + y)(x2 + y2 - xy) we get,

⇒(x)a3+b3.(x)b3+c3.(x)c3+a3⇒xa3+b3+b3+c3+c3+a3⇒x2(a3+b3+c3).\Rightarrow (x)^{a^3 + b^3}.(x)^{b^3 + c^3}.(x)^{c^3 + a^3} \\[1em] \Rightarrow x^{a^3 + b^3 + b^3 + c^3 + c^3 + a^3} \\[1em] \Rightarrow x^{2(a^3 + b^3 + c^3)}.

Hence, (xax−b)a2−ab+b2.(xbx−c)b2−bc+c2.(xcx−a)c2−ca+a2=x2(a3+b3+c3)\Big(\dfrac{x^a}{x^{-b}}\Big)^{a^2 - ab + b^2}.\Big(\dfrac{x^b}{x^{-c}}\Big)^{b^2 - bc + c^2}.\Big(\dfrac{x^c}{x^{-a}}\Big)^{c^2 - ca + a^2} = x^{2(a^3 + b^3+ c^3)}.

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