Given that
y is inversely proportional to the cube of x.
Explanation -
Since it is given that we have to check that for y = 6 and x = 2, the value of variation is 48 or not.
The question can be written as
[tex]y\propto\frac{1}{x^3}[/tex]It can be written as
[tex]\begin{gathered} y=k\times\frac{1}{x^3} \\ where\text{ k = constant of variation} \end{gathered}[/tex]Sustituting the values of x and y.
