Computation of the univariate resultant of two cyclotomic polynomials
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$\mathrm{p1}\u2254{\mathrm{NumberTheory}}_{\mathrm{CyclotomicPolynomial}}\left(1501\,x\right)\:$

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$\mathrm{p2}\u2254{\mathrm{NumberTheory}}_{\mathrm{CyclotomicPolynomial}}\left(1502\,x\right)\:$

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$\mathrm{t1}\u2254\mathrm{time}\left(\mathrm{resultant}\left(\mathrm{p1}\,\mathrm{p2}\,x\right)\right)$

${\mathrm{t1}}{\u2254}{0.005}$
 (1) 
Without the probabilistic approach, this is much more expensive
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$\mathrm{forget}\left(\mathrm{resultant}\right)\:$

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$\mathrm{\_EnvProbabilistic}\u22540\:$

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$\mathrm{t2}\u2254\mathrm{time}\left(\mathrm{resultant}\left(\mathrm{p1}\,\mathrm{p2}\,x\right)\right)$

${\mathrm{t2}}{\u2254}{0.378}$
 (2) 
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$\frac{\mathrm{t2}}{\mathrm{t1}}$

and in fact computation of the resultant bound alone requires more than 10 times the total time required to compute the probabilistic answer on this extreme example.