# Exercise ID276

Algebra → Exponents → Powers
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 11] [Number of characters: 0]

Simplify the following expression:

$\frac{\frac{5{a}^{-x}+5}{{a}^{-2x}-{a}^{-x}+1}}{\frac{1}{{a}^{-x}+1}+\frac{3{a}^{-x}}{{a}^{3x}+1}}$

$\frac{\frac{5{a}^{-x}+5}{{a}^{-2x}-{a}^{-x}+1}}{\frac{1}{{a}^{-x}+1}+\frac{3{a}^{-x}}{{a}^{3x}+1}}$ $=\frac{\frac{\frac{5}{{a}^{x}}+5}{\frac{1}{{a}^{2x}}-\frac{1}{{a}^{x}}+1}}{\frac{1}{{a}^{-x}+1}+\frac{3{a}^{-x}}{{a}^{3x}+1}}$ $=\frac{\frac{\frac{5}{{a}^{x}}+\frac{5}{1}}{\frac{1}{{a}^{2x}}-\frac{1}{{a}^{x}}+1}}{\frac{1}{\frac{1}{{a}^{x}}+1}+\frac{\frac{3}{{a}^{x}}}{{a}^{3x}+1}}$ $=\frac{\frac{\frac{5}{t}+\frac{5}{1}}{\frac{1}{{t}^{2}}-\frac{1}{t}+1}}{\frac{1}{\frac{1}{t}+1}+\frac{\frac{3}{t}}{{t}^{3}+1}}$ $=\frac{\frac{\frac{5+5t}{t}}{\frac{1-t+{t}^{2}}{{t}^{2}}}}{\frac{1}{\frac{1+t}{t}}+\frac{\frac{3}{t}}{\frac{1+{t}^{3}}{{t}^{3}}}}$ $=\frac{\frac{5+5t}{\frac{1-t+{t}^{2}}{t}}}{\frac{1}{\frac{1+t}{t}}+\frac{3}{\frac{1+{t}^{3}}{{t}^{2}}}}$ $=\frac{\frac{5t\left(t+1\right)}{{t}^{2}-t+1}}{\frac{t}{1+t}+\frac{3{t}^{2}}{1+{t}^{3}}}$ $=\frac{\frac{5t\left(t+1\right)}{{t}^{2}-t+1}}{\frac{t}{1+t}+\frac{3{t}^{2}}{\left(1+t\right)\left({t}^{2}-t+1\right)}}$ $=\frac{\frac{5{t}\left(t+1\right)}{{{t}}^{2}{-}{t}{+}{1}}}{\frac{{t}\left({t}^{2}+2t+1\right)}{\left(1+t\right)\left({t}^{2}-t+1\right)}}$ $=\frac{5\left({t}^{2}+2t+1\right)}{{{t}}^{2}{+}{2}{t}{+}{1}}$ $=5$
$\frac{\frac{5{a}^{-x}+5}{{a}^{-2x}-{a}^{-x}+1}}{\frac{1}{{a}^{-x}+1}+\frac{3{a}^{-x}}{{a}^{3x}+1}}=5$