Find the smallest number by which \[243\] must be multiplied to obtain a perfect cube.

Hint:
First, we need to find the prime factor of the given number, which is the number when multiplied together, gives the original number. Prime factor numbers are divisible by \[1\] and by itself. Here, we will multiply the factors with the smallest number to make it a perfect cube, which is equal to the number when raised to the power of three. A perfect cube is checked by grouping the factors of a number grouped into triplets.

Complete step by step solution:
First, find the prime factor of the given number i.e.
\[\left( {243} \right) = 3 \times 3 \times 3 \times 3 \times 3\]
Now, we can see if we multiply the factors \[3 \times 3 \times 3 \times 3 \times 3\] together, we will get the original number \[243\]
Now, we will group the prime factors into triplets since we are looking for the perfect cube
\[\left( {243} \right) = \left[ {\underline {3 \times 3 \times 3} } \right] \times 3 \times 3\]
After grouping the factors into triplets, we can find that there are two 3s left that do not make triplets; hence, we need \[1\] more \[3\] multiplied to the factor to make triplets for a perfect cube.
Now, we multiply the factors with 3 to make it a perfect cube
\[243 \times 3 = \underline {3 \times 3 \times 3} \times \underline {3 \times 3 \times 3} = 729\]
729 is a perfect cube of 9.
Hence, the smallest number which is multiplied with \[243\] to make it a perfect cube is \[3\].

Note:
Generally, the grouping of the factor depends on what is asked by the question of whether its perfect square or perfect cube. In the case of the perfect square, the factors are grouped in pairs, whereas in the case of a cube, factors are grouped in triplets.