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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Updated on: 10-Oct-2022

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