Getting smallest positive number

Question

How can I get the smallest positive number $p$ such that $10500p$ is a perfect cube?

I have tried to express $10500$ into its prime factors but not getting correct solution. Please help

Answer 1

Calling $n=10500$ we need $m$ such that $$n p = m^3$$

now considering the prime factorization $m = \Pi_{k=1}^{\Phi} a^{\alpha_k}$

then $m^3 = \Pi_{k=1}^{\Phi} a^{3\alpha_k} = \Pi_{j=1}^{\Psi}b^{\beta_j} = n p$

Here $\Pi_{j=1}^{\Psi}b^{\beta_j}$ is the prime factorization for $n p$

but here $\Pi_{k=1}^{\Phi} a^{3\alpha_k} =\Pi_{j=1}^{\Psi}b^{\beta_j} = 2^2 \times 3 \times 5^3 \times 7 \times p$

hence $p = 2 \times 3^2 \times 7^2 = 882$

Answer 2

Since $10500 = 2^2 \times 3 \times 5^3 \times 7$, to make $10500p$ a perfect cube, the smallest possible $p$ is $2 \times 3^2 \times 7^2 = 882$, so that $$10500p = (2 \times 3 \times 5 \times 7)^3 = 210^3.$$