We need to evaluate the expression $\sqrt[3]{1372} \times \sqrt[3]{1458}$.
Using the property of radicals, $\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}$, we can combine the terms under a single cube root.
$\sqrt[3]{1372 \times 1458}$.
To simplify this, we find the prime factorization of each number.
For 1372:
$1372 = 2 \times 686 = 2 \times 2 \times 343 = 2^2 \times 7^3$.
For 1458:
$1458 = 2 \times 729 = 2 \times 9^3 = 2 \times (3^2)^3 = 2 \times 3^6$.
Now, multiply the prime factorizations:
$1372 \times 1458 = (2^2 \times 7^3) \times (2 \times 3^6)$.
Combine the powers of the same bases:
$= 2^{2+1} \times 3^6 \times 7^3 = 2^3 \times 3^6 \times 7^3$.
Now, take the cube root of this product:
$\sqrt[3]{2^3 \times 3^6 \times 7^3}$.
$= \sqrt[3]{2^3} \times \sqrt[3]{3^6} \times \sqrt[3]{7^3}$.
This simplifies to $2^{3/3} \times 3^{6/3} \times 7^{3/3}$.
$= 2^1 \times 3^2 \times 7^1$.
$= 2 \times 9 \times 7$.
$= 18 \times 7 = 126$.