Express 1728 as a product of its prime factors in index notation. Using your answer in part (a), explain why 1728 is a perfect cube. k is a prime number. Find the value of k such that $\dfrac{1728}{k}$ is a perfect square.

1. Divide 1728 repeatedly by prime factors and collect the factors.

$1728=2\times 864=2\times2\times432=2^3\times216=2^4\times108=2^5\times54=2^6\times27$
Since $27=3^3$,

2. Write the prime factorisation in index notation.

$1728=2^6\times3^3$

1. Use the prime factorisation to show that the powers can be grouped in threes.

From part (a),
$1728=2^6\times 3^3.$
Since both indices, $6$ and $3$, are multiples of $3$,
$1728=(2^2\times 3)^3=12^3.$
Therefore, $1728$ is a perfect cube.

1. Use the prime factorisation and choose the prime factor that makes all indices even.

From part (a),
$1728=2^6\times 3^3.$
So
$\frac{1728}{k}=\frac{2^6\times 3^3}{k}.$
For this to be a perfect square, all powers must be even. Since $2^6$ already has an even power and $3^3$ has an odd power, divide by $3$:
$\frac{1728}{3}=2^6\times 3^2,$
which is a perfect square.

Hence, $k=3$.