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A root of 10x3−3x2−3x−1=010x^3-3x^2-3x-1=010x3−3x2−3x−1=0 is of the form a3+b3+1c\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}c3a+3b+1, where aaa, bbb, and ccc are positive integers. Find a+b+ca+b+ca+b+c.
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