$T(n)={15n}^3+n^2+4 \\
{15n}^3+n^2+4 \le {15n}^3+n^3+{4n}^3, when \; x>1\\
|{15n}^3+n^2+4| \le |{20n}^3| \\
Thus \; c=20, n_0=1$
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