Not too tough

Using the properties of exponents the expression can be written as: $\ ( x^{a-b} )^{a+b} \times\ ( x^{b-c} )^{b+c} \times\ ( x^{c-a} )^{c+a}$ $\ = x^{a^2 -b^2} . x^{b^2 -c^2} . x^{c^2 -a^2}$ $\ = x^ {(a^2 -b^2) + (b^2 -c^2) + (c^2 -a^2)} = x^0 = 1 \ (x > 0)$ Also, note that the exponent of x has to be zero for the expression to be a constant (as implied by the question). This can be proven with logs... $\ x^{\phi} = C \implies\ \phi log (x) = log (C)$ Now log(x) varies as x varies, so we must have $\phi$ = 0 for the LHS to be equal to a constant, given as log(C). $\therefore\ x^{\phi} = x^0 = 1$