{
    "problem": "Find\n\\[\\cos \\left( 6 \\arccos \\frac{1}{3} \\right).\\]",
    "level": "Level 3",
    "type": "Precalculus",
    "solution": "Let $x = \\arccos \\frac{1}{3},$ so $\\cos x = \\frac{1}{3}.$  From the triple angle formula,\n\\[\\cos 3x = 4 \\cos^3 x - 3 \\cos x = 4 \\left( \\frac{1}{3} \\right)^3 - 3 \\cdot \\frac{1}{3} = -\\frac{23}{27}.\\]Then from the double angle formula,\n\\[\\cos 6x = 2 \\cos^2 3x - 1 = 2 \\left( -\\frac{23}{27} \\right)^2 - 1 = \\boxed{\\frac{329}{729}}.\\]"
}