Question

For any positive real number x, find the value of \[\left( \frac{x^a}{x^b} \right)^{a + b} \times \left( \frac{x^b}{x^c} \right)^{b + c} \times \left( \frac{x^c}{x^a} \right)^{c + a}\].

Answer 1

We have to find the value of L =   `(x^a/x^a)^(a+b) xx (x^b/x^c)^(b+c) xx(x^c/x^a)^(c+a) `

 `L=(x^(a(a+b))/(x^(b(a+b)))) xx (x^(b(b+c))/(x^(c(b+c)))) xx  (x^(c(c+b))/(x^(a(c+a)))) ` 

`= (x^(a^2+ab))/ (x^(ba+b^2)) xx (x^(b^2+bc))/ (x^(bc+c^2)) xx (x^(c^2+ca))/ (x^(ac+b^2))`

By using rational exponents, `a^mxx a^n = a^(m+n)` we get

  `L=(x^(a^2+ab+b^2+bc+c^2+ca))/(x^(ab+b^2+bc+c^2+ac+a^2))`

By using rational exponents    `a^m/a^n= a^(m-n)` we get 

`L = x^((a^2+ab+b^2+bc+c^2+ca) -(ab+b^2+bc+c^2+ac+a^2))`

`=x^((a^2+ab+b^2+bc+c^2+ca) -(ab+b^2+bc+c^2+ac+a^2))`

`=x^0`

By definition we can write `x^0` as 1

Hence the value of expression is 1.