Question

If the mean of $x$ and $\dfrac{1}{x}$ is M, then the mean of ${x^3}$ and $\dfrac{1}{{{x^3}}}$ is
A. $M({M^2} + 3)$
B. $M(4{M^2} - 3)$
C. ${M^3}$
D. ${M^3} + 3$

Answer 1

Hint: According to given in the question we have to determine the mean of ${x^3}$ and $\dfrac{1}{{{x^3}}}$ if the mean of $x$ and $\dfrac{1}{x}$ is M. So, first of all we have to understand about the mean which is as explained below:
Mean: The mean or we can say that the average of a data set which can be found by adding all the given numbers in the data set and then dividing by the number of the values in the set.
Now, we have to find the mean of both of the terms as $x$ and $\dfrac{1}{x}$ so, first of all we have to find the sum of both of the term $x$ and $\dfrac{1}{x}$ then we have to divide it with the total number of terms we added which are 2 in total to determine the mean.

Formula used:
$ {(a + b)^3} = {a^3} + {b^3} + 3ab(a + b)..........................(A)$

Complete step-by-step solution:
Step 1: First of all we have to determine the sum of both of the terms as given $x$ and $\dfrac{1}{x}$ mean that we have already understood the solution hint. Hence,
$ = x + \dfrac{1}{x}$
Step 2: Now, as we know that the total number of terms are 2 which are $x$and $\dfrac{1}{x}$so, we can determine the mean we just have to divide the obtained sum with the total number of terms as mentioned in the solution hint.
$ \Rightarrow $Mean$ = \dfrac{{x + \dfrac{1}{x}}}{2}$……………….(1)
Step 3: Now, as mentioned in the question, the mean of both of the terms is M on substituting these values in the expression (1) as obtained in the solution step 2. Hence,
$
   \Rightarrow \dfrac{{x + \dfrac{1}{x}}}{2} = M \\
   \Rightarrow x + \dfrac{1}{x} = 2M.............(2)
 $
Step 4: Now, we have to find the cube of the both sides of the expression as obtained in the solution step 3.
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^3} = {(2M)^3}$…………….(3)
Step 5: Now, to find the cube root of the expression as obtained in the solution step 4 we have to use the formula (A) as mentioned in the solution hint. Hence,
\[
   \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3 \times x \times \dfrac{1}{x}\left( {x + \dfrac{1}{x}} \right) = 8{M^3} \\
   \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 3\left( {x + \dfrac{1}{x}} \right) = 8{M^3}
 \]
Now, as we know that \[\left( {x + \dfrac{1}{x}} \right) = 2M\] hence, on substituting this in the expression as obtained just above,
\[
   \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 6M = 8{M^3} \\
   \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 8{M^3} - 6M
 \]
Now, we have to take M as a common term from the right side of the expression as obtained just above,
\[ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = M(8{M^2} - 6)\]
Step 6: Now, we have to find the mean of the terms ${x^3}$ and $\dfrac{1}{{{x^3}}}$, the same as step 3.
$ \Rightarrow $Mean:
$
   = \dfrac{{{x^3} + \dfrac{1}{{{x^3}}}}}{2} \\
   = \dfrac{{M(8{M^2} - 6)}}{2} \\
   = M(4{M^2} - 3)
 $
Hence, with the help of the formula (A) as mentioned in the solution hint we have determined the value of mean for the given terms ${x^3}$ and $\dfrac{1}{{{x^3}}}$ is $M(4{M^2} - 3)$.

Therefore option (B) is correct.

Note: To determine the mean for the given terms ${x^3}$ and $\dfrac{1}{{{x^3}}}$ it is necessary that we have to determine the mean for the terms $x$ and $\dfrac{1}{x}$ whose mean is M and on solving the mean and applying the cube in the both sides of the equation obtained we can determine the value of ${x^3}$ and $\dfrac{1}{{{x^3}}}$.
Mean is basically the same as the average for the given data set in which first of all we have to determine the sum of all the given numbers or data and then we have to divide the sum by the total number of data sets.