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Simplify: (xa+b)ab.(xb+c)bc.(xc+a)ca(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a}

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(xa+b)ab.(xb+c)bc.(xc+a)ca=(x)(a+b)×(ab).(x)(b+c)×(bc).(x)(c+a)×(ca)=(x)a2b2.(x)b2c2.(x)c2a2=(x)a2b2+b2c2+c2a2=(x)0=1(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a}\\[1em] = (x)^{(a+b)\times(a-b)}.(x)^{(b+c)\times(b-c)}.(x)^{(c+a)\times(c-a)}\\[1em] = (x)^{a^2-b^2}.(x)^{b^2-c^2}.(x)^{c^2-a^2}\\[1em] = (x)^{a^2-b^2+b^2-c^2+c^2-a^2}\\[1em] = (x)^{0}\\[1em] = 1

(xa+b)ab.(xb+c)bc.(xc+a)ca=1(x^{a+b})^{a-b}.(x^{b+c})^{b-c}.(x^{c+a})^{c-a} = 1

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