Let \(z = 6^x\).
Since \(x + y = 3\), \[
6^y = 6^{3 - x} = \frac{6^3}{6^x} = \frac{36 \times 6}{z}
\]
Hence\[
6^x + 6^y = z + \frac{36 \times 6}{z} = 42
\]
We have\[
z^2 - 42 z + 36 \times 6 = 0 \Rightarrow (z - 36)(z - 6) = 0
\]
There are two solutions \(z = 36\) and \(z = 6\) which implies \(x = 2, y = 1\) and \(x = 1, y = 2\).
Since \(x + y = 3\), \[
6^y = 6^{3 - x} = \frac{6^3}{6^x} = \frac{36 \times 6}{z}
\]
Hence\[
6^x + 6^y = z + \frac{36 \times 6}{z} = 42
\]
We have\[
z^2 - 42 z + 36 \times 6 = 0 \Rightarrow (z - 36)(z - 6) = 0
\]
There are two solutions \(z = 36\) and \(z = 6\) which implies \(x = 2, y = 1\) and \(x = 1, y = 2\).
