If x varies inversely as $$y^3$$​ - 1 and is equal to 3 when y = 4, find x when y = 6.

Correct option is B

Given:

x varies inversely as $$y^3 - 1$$​, i.e.,

​$$x \propto \frac{1}{y^3 - 1}.$$​​

x = 3 when y = 4.

We need to find x when y = 6.

Solution:

The relationship between xx and $$y^3 - 1$$​ can be expressed as:

​$$x = \frac{k}{y^3 - 1}$$​​

where k is a constant.

Using the given values x = 3 and y = 4:

​$$3 = \frac{k}{4^3 - 1} = \frac{k}{64 - 1} = \frac{k}{63}$$​​

​$$k = 3 \times 63 = 189$$​​

Now, x when y = 6:

​$$x = \frac{189}{6^3 - 1} = \frac{189}{216 - 1} = \frac{189}{215}$$​​