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Naily [24]
2 years ago
8

%20%7Bx%7D%5E%7Bc%20%2B%20a%7D%20%7D%7B%20%7Bx%7D%5E%7Ba%7D.%20%7Bx%7D%5E%7Bb%7D%20%20.%20%7Bx%7D%5E%7Bc%7D%20%7D%20%20%5C%5C%20" id="TexFormula1" title="Simplify: \frac{ {x}^{a + b} . {x}^{b + c} {x}^{c + a} }{ {x}^{a}. {x}^{b} . {x}^{c} } \\ " alt="Simplify: \frac{ {x}^{a + b} . {x}^{b + c} {x}^{c + a} }{ {x}^{a}. {x}^{b} . {x}^{c} } \\ " align="absmiddle" class="latex-formula">
Need help please.​
Mathematics
1 answer:
kozerog [31]2 years ago
7 0

Step-by-step explanation:

\bf \underline{Solution-} \\

We have to simplify the given expression.

\rm =  \dfrac{ {x}^{a + b} \cdot {x}^{b + c} \cdot {x}^{c + a}  }{ {x}^{a} \cdot {x}^{b} \cdot {x}^{c} }

We know that:

\rm \longmapsto {x}^{a} \times  {x}^{b}  =  {x}^{a + b}

\rm \longmapsto \dfrac{ {x}^{a} }{ {x}^{b} } =  {x}^{a -  b}

Therefore, we get:

\rm =  \dfrac{ {x}^{(a + b) + (b + c) + (c + a)}}{ {x}^{a + b + c} }

\rm =  \dfrac{ {x}^{2(a + b+ c)}}{ {x}^{a + b + c} }

\rm = {x}^{2(a + b+ c) - (a + b + c)}

\rm = {x}^{a + b + c}

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To know about standard deviation click here :

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