### Instruction: Find
\[\cos \left( 6 \arccos \frac{1}{3} \right).\].### Response: This problem involves an inverse trigonometric function and a multiple angle formula.
To simplify the expression, I will use the fact that \[\cos (2x) = 2 \cos^2 x - 1.\]
This means that I can write \[\cos \left( 6 \arccos \frac{1}{3} \right) = \cos \left( 2 \cdot 3 \arccos \frac{1}{3} \right) = 2 \cos^2 \left( 3 \arccos \frac{1}{3} \right) - 1.\]
Now I need to find the value of \[\cos \left( 3 \arccos \frac{1}{3} \right).\]
I can use another multiple angle formula, which says that \[\cos (3x) = 4 \cos^3 x - 3 \cos x.\]
This means that \[\cos \left( 3 \arccos \frac{1}{3} \right) = 4 \left( \cos \left( \arccos \frac{1}{3} \right) \right)^3 - 3 \cos \left( \arccos \frac{1}{3} \right).\]
But I know that \[\cos \left( \arccos \frac{1}{3} \right) = \frac{1}{3},\] because that is how the inverse cosine function works.
So I can substitute \[\cos \left( 3 \arccos \frac{1}{3} \right) = 4 \left( \frac{1}{3} \right)^3 - 3 \left( \frac{1}{3} \right) = \frac{4}{27} - 1 = -\frac{23}{27}.\]