We'll rewrite the expression using the following properties:
[tex] x^{ a^{b} }=x^{ab} [/tex]
[tex] x^{a} x^{b} = x^{a+b} [/tex]
[tex] \frac{ x^{a} }{x^{b}}= x^{a-b} [/tex]
[tex] (a+b)^{2}=a^2+2ab+b^2 [/tex]
[tex](a+b)(a-b)=a^2-b^2[/tex]
In which [tex]a[/tex] and [tex]b[/tex] are any real numbers.
So,
[tex] \frac{x^a}{x^b}(a^2+ab+b^2)* \frac{x^b}{x^c}(b^2+bc+c^2)* \frac{x^c}{x^a}(a^2+ac+c^2)[/tex]
By applying the previous properties can be rewritten as:
[tex] x^{(a-b)[ (a+b)^{2} -ab]}* x^{(b-c)[ (b+c)^{2} -bc]}* x^{(c-a)[ (a+c)^{2} -ac]} [/tex]
We keep rewriting:
[tex]
x^{(a-b)[ (a+b)^{2} -ab]+(b-c)[ (b+c)^{2} -bc]+(c-a)[ (a+c)^{2} -ac]}
[/tex]
So, we need to prove the previous is equal to one.
We recall the property of exponential:
[tex] x^{0}=1 [/tex]
So, proving the following suffices:
[tex](a-b)[ (a+b)^{2} -ab]+(b-c)[ (b+c)^{2} -bc]+(c-a)[ (a+c)^{2} -ac]=0[/tex]
Proving the previous is quite simple, you can just computed directly, and you'll see that eventually everything cancels out. We can also rewrite first the expression in a more "digestible" form, as follows:
[tex](a-b)[ (a+b)^{2} -ab]+(b-c)[ (b+c)^{2} -bc]+(c-a)[ (a+c)^{2} -ac]=[/tex]
[tex](a^2-b^2)(a+b)+ab^2-a^2b+(b^2-c^2)(b+c)+bc^2-b^2c+
[/tex]
[tex](c^2-a^2)(c+a)+a^2c-ac^2[/tex]
Which you can very easily shown to be equal to 0.
