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(3^(4n)-3^(2n))\ (3^(3n) + 3^(2n))

 Jul 29, 2015

Best Answer 

 #1
avatar+26367 
+13

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } \\
&=& \dfrac{ 3^{2n+2n}-3^{2n} } { 3^{2n+n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n}3^{2n}-3^{2n} } { 3^{2n}3^{n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n} (3^{2n}-1) } { 3^{2n} (3^{n} + 1) } \\\\
&=& \dfrac{ 3^{2n}-1 } { 3^{n}+1 } \\\\
&=& \dfrac{ (3^{n}-1)(3^{n}+1) } { 3^{n}+1 } \\\\
\mathbf{ \dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } }&\mathbf{=} & \mathbf{3^{n}-1}
\end{array}
$}}$$

 

.
 Jul 29, 2015
 #1
avatar+26367 
+13
Best Answer

$$\small{\text{$
\begin{array}{rcl}
\dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } \\
&=& \dfrac{ 3^{2n+2n}-3^{2n} } { 3^{2n+n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n}3^{2n}-3^{2n} } { 3^{2n}3^{n} + 3^{2n} } \\\\
&=& \dfrac{ 3^{2n} (3^{2n}-1) } { 3^{2n} (3^{n} + 1) } \\\\
&=& \dfrac{ 3^{2n}-1 } { 3^{n}+1 } \\\\
&=& \dfrac{ (3^{n}-1)(3^{n}+1) } { 3^{n}+1 } \\\\
\mathbf{ \dfrac{ 3^{4n}-3^{2n} } { 3^{3n} + 3^{2n} } }&\mathbf{=} & \mathbf{3^{n}-1}
\end{array}
$}}$$

 

heureka Jul 29, 2015

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