Concept:For 0≤x≤1, use the identity sin−1x+cos−1x=2π and the expansion of p3+q3.Explanation:Let sin−1x=θ, so cos−1x=2π−θ, where 0≤θ≤2π.Then the given expression becomes θ3+(2π−θ)3.Using p3+q3=(p+q)3−3pq(p+q), we get:θ3+(2π−θ)3=(2π)3−3θ(2π−θ)(2π).Since θ and 2π−θ are non-negative with fixed sum 2π, their product is maximum when they are equal.Thus, θ(2π−θ)≤(4π)2.Hence, the expression satisfies:θ3+(2π−θ)3≥8π3−3(16π2)(2π)=32π3.Given (sin−1x)3+(cos−1x)3=aπ3, we have aπ3≥32π3.Therefore, a≥321.Answer:Option A: a≥321.