$\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}=?$
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3
9
27
Explanation:
Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$
$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$
$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$
$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$
$=3^{\left(3n+1-3n+1\right)}$
$=3^{2}=9$
Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$
$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$
$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$
$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$
$=3^{\left(3n+1-3n+1\right)}$
$=3^{2}=9$