Verified answer

Answer:

Consider,

[tex]\sf \:\dfrac{1}{ {x}^{a - b} + 1} + \dfrac{1}{ {x}^{b - a} + 1} \\ \\ [/tex]

can be rewritten as

[tex]\qquad\sf \: = \: \dfrac{1}{ \dfrac{ {x}^{a} }{ {x}^{b} } + 1} + \dfrac{1}{\dfrac{ {x}^{b} }{ {x}^{a} } + 1} \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: {x}^{m - n} = \dfrac{ {x}^{m} }{ {x}^{n} } \: }} \\ \\ [/tex]

[tex]\qquad\sf \: = \: \dfrac{1}{ \dfrac{ {x}^{a} + {x}^{b} }{ {x}^{b} }} + \dfrac{1}{\dfrac{ {x}^{b} + {x}^{a} }{ {x}^{a} }} \\ \\ [/tex]

[tex]\qquad\sf \: = \:\dfrac{ {x}^{b} }{ {x}^{a} + {x}^{b} } + \dfrac{ {x}^{a} }{ {x}^{b} + {x}^{a} } \\ \\ [/tex]

[tex]\qquad\sf \: = \:\dfrac{ {x}^{b} }{ {x}^{b} + {x}^{a} } + \dfrac{ {x}^{a} }{ {x}^{b} + {x}^{a} } \\ \\ [/tex]

[tex]\qquad\sf \: = \:\dfrac{ {x}^{b} + {x}^{a} }{ {x}^{b} + {x}^{a} } \\ \\ [/tex]

[tex]\qquad\sf \: = \:1\\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \:\dfrac{1}{ {x}^{a - b} + 1} + \dfrac{1}{ {x}^{b - a} + 1} = 1 \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]