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If $ 3^{3000}-3^{2999}-3^{2998}-3^{2997}=a \cdot 3^{2997} $, find the value of $ a $.
Given:
\( 3^{3000}-3^{2999}-3^{2998}-3^{2997}=a \cdot 3^{2997} \)
To do:
We have to find the value of \( a \).
Solution:
$ \begin{array}{l}
3^{3000} -3^{2999} -3^{2998} -3^{2997} =3^{2997}\left( 3^{3} -3^{2} -3^{1} -1\right)\\
=3^{2997}( 27-9-3-1)\\
=3^{2997}( 27-13)\\
=14\ .\ 3^{2997}
\end{array}$
$=a \cdot 3^{2997}$
Comparing both sides, we get,
$a=14$.
The value of $a$ is $14$.
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