Question

Prove the following:

`(x^("a"+"b")/x^"c")^("a"-"b") · (x^("c"+"a")/(x^"b"))^("c"-"a") · ((x^("b"+"c"))/(x"a"))^("b"-"c")` = 1

Answer 1

L.H.S.

= `("a"^"m"/"a"^"n")^("m"+"n"+1) ·("a"^"n"/"a"^1)^("n" + 1-"m").("a"^1/"a"^"m")^(1+"m"-"n")`

= `"a"^("m"("m" + "n" - 1))/("a"^("n"("m"+"n" - 1)))·"a"^("n"("n" + 1 - "m"))/("a"^(1("n"+ 1 - "m")))·"a"^(1(1 + "m" - "n")) /"a"^("m"(1 + "m" - "n")) `      ......(Using(a^m)ⁿ = a^mn)

= `"a"^("m"^z + "mn" - "m")/"a"^("n"^z+"mn" - "n")·"a"^("n"^z - "mn" + "n")/"a"^("n"+1-"m")·"a"^(1 + "m" - "n")/"a"^("m"^z - "mn" + "m")`

= `"a"^("m"^z + "mn" - "m" - ("n"^z+"mn"-"n")) ·"a"^("n"^z - "mn" - ("n" + 1 - "m"))·"a"^(1+"m"-"n"-("m"^z-"mn"+"m"))`   ....(Using a^m ÷ aⁿ = a^m-n)

= `"a"^("m"^z+"mn"-"m"-"n"^z-"mn"+"n")·"a"^("n"^z-"mn"+"n"-"n"-1+"m")·"a"^(1+"m"-"n"-"m"^z-"mn"+"m")`

= `"a"^("m"^z + "mn"-"m"-"n"^z-"mn"+"n"+"n"^z-"mn"+"n"-"n"-1+"m"+1+"m"-"n"-"m"^z+"mn"-"m")`   ....(Using a^m x aⁿ = a^m+n)

= a°
= 1    .....(Using a° = 1)
= R.H.S.
Hence proved.