Question

Simplify the following :


`\begin{gathered} \: \boxed{\sf \: {\bigg( {x}^{ (b)/(b - c) } \bigg) }^{ (1)/(b - a) } * {\bigg( {x}^{ (c)/(c - a) } \bigg) }^{ (1)/(c - b) } * {\bigg( {x}^{ (a)/(a - b) } \bigg) }^{ (1)/(a - c) }} \end{gathered}`

Answer 1

1

Explanation:

`\left(x^{\tfrac b{b-c}} \right)^{\tfrac 1{b-a}} * \left(x^{\tfrac c{c-a}} \right)^{\tfrac 1{c-b}} * \left(x^{\tfrac a{a-b}} \right)^{\tfrac 1{a-c}}\\\\\\=x^{\tfrac{b}{(b-c)(b-a)}} \cdot x^{\tfrac{c}{(c-a)(c-b)}} \cdot x^{\tfrac{a}{(a-b)(a-c)}}~~~~~;\left[\text{Apply exponent rule:}~ (a^m)^n = a^(mn) \right]\\\\=x^{\tfrac{b}{(b-c)(b-a)} + \tfrac{c}{(c-a)(c-b)} + \tfrac{a}{(a-b)(a-c)}} ~~~~~~~~;\left[\text{Apply exponent rule:}~ a^m \cdot a^n = a^(m+n) \right]\\\\`

`=x^{\tfrac{b}{-(c-b)(b-a)} + \tfrac{c}{-(a-c)(c-b)} + \tfrac{a}{-(b-a)(a-c)}} \\\\=x^{-\tfrac{b}{(c-b)(b-a)} - \tfrac{c}{(a-c)(c-b)} -\tfrac{a}{(b-a)(a-c)}} \\\\=x^{-\left[\tfrac{b(a-c)+c(b-a)+a(c-b)}{(c-b)(b-a)(a-c)} \right]}\\\\=x^{-\left[\tfrac{ba-bc+cb-ac+ac-ab}{(c-b)(b-a)(a-c)} \right]}\\\\=x^{\tfrac{-ba+bc-cb+ac-ac+ab}{(c-b)(b-a)(a-c)} }\\\\=x^{\tfrac{0}{(c-b)(b-a)(a-c)}}\\\\=x^0\\\\=1`