Express 378 as the product of its prime factors. Answer ... Find the smallest positive integer $k$ such that 378$k$ is a perfect cube. Answer $k =$ ...

1. Divide 378 successively by prime factors until only primes remain, then write the product in index form.

$378 = 2 \times 189$

$189 = 3 \times 63 = 3 \times 3 \times 21 = 3 \times 3 \times 3 \times 7$

So,
$378 = 2 \times 3 \times 3 \times 3 \times 7 = 2 \times 3^3 \times 7.$

1. Use the prime factorisation of 378 and make all prime exponents multiples of 3.

From part (a),
$378=2\times 3^3\times 7.$
For $378k$ to be a perfect cube, the exponents of all prime factors must be multiples of $3$.

$2^1$ needs another $2^2$, and $7^1$ needs another $7^2$.

So
$k=2^2\times 7^2=4\times 49=196.$