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algebra intermediate

Problem

The fraction

\frac{( 3 ^{2008} ) ^{2} - ( 3 ^{2006} ) ^{2}}{( 3 ^{2007} ) ^{2} - ( 3 ^{2005} ) ^{2}}

simplifies to which of the following?

(A)

1

(B)

\frac{9}{4}

(C)

3

(D)

\frac{9}{2}

(E)

9

Solution

Using Difference of Squares,

\frac{( 3 ^{2008} ) ^{2} - ( 3 ^{2006} ) ^{2}}{( 3 ^{2007} ) ^{2} - ( 3 ^{2005} ) ^{2}}

becomes

\frac{( 3 ^{2008} + 3 ^{2006} ) ( 3 ^{2008} - 3 ^{2006} )}{( 3 ^{2007} + 3 ^{2005} ) ( 3 ^{2007} - 3 ^{2005} )}

= \frac{3 ^{2006} ( 9 + 1 ) \cdot 3 ^{2006} ( 9 - 1 )}{3 ^{2005} ( 9 + 1 ) \cdot 3 ^{2005} ( 9 - 1 )}

= 9

Final answer

E