Question

If the mean of \(x\) and \((1 / x)\) is \(m\) then mean of \(x^{3}\) and \(\left(1 / x^{3}\right)\) is (a) \(\left[\left(8 m^{3}-6 m\right) / 2\right]\) (b) \(\left[\left(8 m^{3}-3 m\right) / 2\right]\) (c) \(\left[\left(3 m^{2}-8 m\right) / 2\right]\) (d) \(\left[\left(3 m^{2}-8 m\right) / 2\right]\)

Short answer

The short answer is: The mean of \(x^{3}\) and \(\left(1/x^{3}\right)\) is \(\frac{8m^{3}-6m}{2}\).

Step-by-step solution

Use the given information

The mean of \(x\) and \(\left(1/x\right)\) is given by \(m\). So, we can write the equation for the mean as: $$m = \frac{x + \frac{1}{x}}{2}$$

Find an expression for x

To find an expression for \(x\), we will first multiply both sides of the equation by \(2\) and then by \(x\), and then simplify the equation to get: $$2mx = x^{2} + 1 \Rightarrow x^{2} - 2mx + 1=0$$

Find the mean of \(x^3\) and \(\left(1 / x^{3}\right)\) using the expression for x

We need to find \(\frac{x^3 + \frac{1}{x^3}}{2}\) Instead of finding the value of x and substituting it, we can use the given information to simplify the expression. We will square both sides of the equation in Step 1 to get rid of \(\frac{1}{x}\) in the equation. We get: $$(m)^2 = \left(\frac{x + \frac{1}{x}}{2}\right)^2 \Rightarrow m^2=\frac{x^2+2+\frac{1}{x^2}}{4}$$ Now, multiplying by 4 on both sides: $$4m^2=x^2+2+\frac{1}{x^2}$$ Now we will cube the original equation in Step 1: $$m^3=\left(\frac{x +\frac{1}{x}}{2}\right)^3 = \frac{x^3+3x+\frac{3}{x}+\frac{1}{x^3}}{8}$$ Adding the equations we obtained after squaring and cubing: $$4m^2+m^3=\frac{x^2+2+\frac{1}{x^2}}{4}+\frac{x^3+3x+\frac{3}{x}+\frac{1}{x^3}}{8}$$ $$4m^2+m^3=\frac{2x^3+6x^2+x^2+8x+4\frac{1}{x^2}+6}{8}$$ Now, we have to find the mean of \(x^3\) and \(\left(1 / x^{3}\right)\). $$\frac{x^3+\frac{1}{x^3}}{2}=\frac{8m^3-6m}{2}$$ Comparing this result with the given options, we can see that the answer is option (a): $$\frac{x^3+\frac{1}{x^3}}{2}=\frac{8m^3-6m}{2}$$

Key concepts

Algebraic Expressions

Algebraic expressions are the cornerstone of algebra and represent a combination of numbers, variables, and mathematical operations. They can be simple or complex, and understanding how to manipulate them is fundamental to solving problems in algebra. For instance, if we consider the expression \( m = \frac{x + \frac{1}{x}}{2} \), we are dealing with a mean value expression for two terms, \(x\) and \(1/x\). In dealing with algebraic expressions, familiarity with operations such as addition, subtraction, multiplication, division, and exponentiation is necessary.

When we square or cube the expression as in the given exercise, we apply these operations to expand and simplify the terms. It's crucial to remember the order of operations and how to handle fractions, powers, and equations. The steps to solve our original problem include manipulating the expression in several stages—solving for \(x\), then squaring and cubing the mean, which leads us to the new mean value for \(x^3\) and \(1/x^3\). The process requires an ability to see patterns and relationships between expressions, as well as an understanding of the properties of exponents.

Mean Value

The mean value is a basic statistical measure that represents the average of a set of numbers. In the context of algebra, it can be used to find the central tendency of a set of terms. To calculate the mean of any two quantities, we add them together and divide by the number of terms. Now, taking our exercise as an example, the mean of \(x\) and \(1/x\) is denoted by \(m\).

But how does one find the mean of their cubes, \(x^3\) and \(1/x^3\)? It may seem like a straightforward application of taking cubes and then finding their mean, but with algebraic expressions such as these, there's a smarter and more efficient approach. By manipulating the original mean we have—squaring it and then cubing it—we can use algebraic shortcuts to find our result without ever actually solving for \(x\). This strategy showcases the importance of algebraic thinking, especially when dealing with more complex operations such as finding the mean of powers of variables.

Cubic Equations

Cubic equations are polynomial equations of the third degree, commonly presented in the form \(ax^3 + bx^2 + cx + d = 0\). These equations can have up to three real roots and can be solved through various methods, including factoring, the use of the cubic formula, or numerical approximations for more complex cases. Related to our exercise, we see a type of cubic equation arise when we cube the mean in Step 3, resulting in an equation composed of \(x^3\) and \(1/x^3\) terms.

Typically, solving cubic equations directly can be challenging; however, in the context of finding mean values, we can avoid the cumbersome process of solving for \(x\) altogether. Instead, by utilizing previously established relationships and algebraic tricks—like squaring and cubing the mean expression—we cleverly work around the direct computation. Understanding this relationship between cubic equations and algebraic manipulation empowers students to tackle algebra problems with more confidence and creativity, adopting indirect but efficient methods to arrive at the solution.