A cube edge \[3\]cm of iron weighs \[12\]g. What is the weight of the similar cube of iron whose edge is \[12\]cm?

Hint:Before, finding the weight we will find the volume of the iron cube. With the help of a unitary method we will find the required weight.
Unitary method: The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value. In essence, this method is used to find the value of a unit from the value of a multiple, and hence the value of a multiple.

Formula used:
Let us consider, the edge of the cube is \[a\]cm. so, its volume will be \[{a^3}\] cubic cm.

Complete step-by-step answer:
It is given that the length of edge of the cube is \[3\]cm.
The weight of this iron is \[12\]g.
We have to find the weight of the iron when the edge of the cube is \[12\]cm.
At first, we will find the volume of the cube when the edge is \[3\]cm.
Let us consider, the edge of the cube is \[a\]cm. so, its volume will be \[{a^3}\]cubic cm.
The volume of the cube when the length of the edge is \[3\]cm is \[{3^3} = 27\]cubic cm
The volume of the cube when the length of the edge is \[12\]cm is \[{12^3} = 1728\]cubic cm
Now we will apply a unitary method.
When the volume of a cube is \[27\]cubic cm, it weighs \[12\]g. Then the volume of a cube is \[1\]cubic cm, it weighs \[\dfrac{{12}}{{27}}\]g.
When the volume of a cube is \[1728\] cubic cm, it weighs
Weight= \[\dfrac{{12}}{{27}} \times 1728\]g….(1)
That is the weight of a cube with volume 1728 cubic cm is just the multiple of 1728 with the weight of 1 cubic cm of a cube
Let us find the weight by solving the equation (1).
The weight of the iron is \[768\]g.
Hence, we have found the weight of the similar cube of iron whose edge is \[12\] cm is \[768\] g.

Note:The weight of the iron cube depends on the volume of the cube. With the change in the volume of the cube the weight of it changes. And volume is basically measured in cubic units. Whereas the weight is measured in grams.