If x varies inversely as y³ - 1 and is equal to 6 when y = 3, find x when y = 8.

Correct option is C

Given:

If x varies inversely as y³ - 1 and is equal to 6 when y = 3, find x when y = 8.

Solution:

We are given that x varies inversely as $$y^3 -1$$​ This means we can express this relationship as:

$$x = \frac{k}{y^3 - 1}$$ 

where k is a constant. We are also told that x=6 when y=3. Using this information, we can find the value of k. 

Substitute x=6 and y=3 into the equation: 

​$$6 = \frac{k}{3^3 - 1}$$ 

First, calculate $$( 3^3 - 1)$$ 

​$$3^3 = 27 \quad \Rightarrow \quad 3^3 - 1 = 27 - 1 = 26$$​​

​Now, substitute this into the equation:

​$$6 = \frac{k}{26}$$​​

k = 6 × 26 = 156

Now that we know k=156, we can use the equation $$x = \frac{k}{y^3 - 1}$$​ to find x when y=8. Substitute k=156 and y=8 into the equation:

​$$x = \frac{156}{8^3 - 1}$$ 

First, calculate $$( 8^3 - 1)$$ 

​$$8^3 = 512 \quad \Rightarrow \quad 8^3 - 1 = 512 - 1 = 511$$ 

​Now, substitute this into the equation:

​$$x = \frac{156}{511}$$ 

​Thus, the value of $x$ when $y=8$ is: $$x = \frac{156}{511}$$ .​

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