Express 84 as a product of prime factors. Answer ___ The sequence $84p$, $84q$, $84r$ are perfect cubes arranged in ascending order. Find the smallest possible integer value of $q$. Answer $q$ = ___

1. Divide 84 successively by prime factors until only primes remain.

$84 = 2 \times 42$

$= 2 \times 2 \times 21$

$= 2 \times 2 \times 3 \times 7$

Therefore,

$84 = 2^2 \times 3 \times 7$

1. Find the smallest multiplier needed to make $84$ a perfect cube.

From part (i),
$84=2^2\times 3\times 7.$
For $84p$ to be a perfect cube, each prime factor must have an exponent that is a multiple of $3$.

So multiply by
$2^1\times 3^2\times 7^2=2\times 9\times 49=882.$
Hence the smallest possible perfect cube of the form $84p$ is
$84\times 882=2^3\times 3^3\times 7^3=42^3.$

2. Choose the second smallest possible multiplier since $84p$, $84q$, $84r$ are in ascending order.

The perfect cubes that are multiples of $84$ are obtained by multiplying the smallest one by cube numbers:
$84(882\times 1^3),\quad 84(882\times 2^3),\quad 84(882\times 3^3),\ldots$
Since $84p,84q,84r$ are three perfect cubes in ascending order, the smallest possible value of $q$ is the second multiplier:
$q=882\times 2^3=882\times 8=7056.$