[ ($2^{2n}$ -3) ($2^{2n-2}$ )( $3^{n}$ -2)( $3^{n-2}$ )] / $3^{n-4}$ ( $4^{n+3}$ - $2^{2n}$ )

asked by guest
on Nov 19, 2024 at 7:14 am



You asked:

Evaluate the expression: \(\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \cdot \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \cdot \left({4}^{n + 3} - {2}^{2 n}\right)}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \cdot \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \cdot \left({4}^{n + 3} - {2}^{2 n}\right)} = \frac{2^{2 n - 2} \cdot 3^{4 - n} 3^{n - 2} \cdot \left(2^{2 n} - 3\right) \left(3^{n} - 2\right)}{- 2^{2 n} + 4^{n + 3}} \)

Expanded

\[\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \cdot \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \cdot \left({4}^{n + 3} - {2}^{2 n}\right)} = \frac{9 \cdot 2^{4 n} 3^{n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} - \frac{18 \cdot 2^{4 n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} - \frac{27 \cdot 2^{2 n} 3^{n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}} + \frac{54 \cdot 2^{2 n}}{- 4 \cdot 2^{2 n} + 256 \cdot 4^{n}}\]

Factored

\[\frac{\left({2}^{2 n} - 3\right) \cdot {2}^{2 n - 2} \cdot \left({3}^{n} - 2\right) \cdot {3}^{n - 2}}{{3}^{n - 4} \cdot \left({4}^{n + 3} - {2}^{2 n}\right)} = - \frac{2^{2 n - 2} \cdot 3^{4 - n} 3^{n - 2} \cdot \left(2^{2 n} - 3\right) \left(3^{n} - 2\right)}{2^{2 n} - 64 \cdot 4^{n}}\]

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