Simplifying Algebraic Expressions A Step By Step Guide

In the realm of mathematics, simplification of complex expressions is a fundamental skill. This article delves into the step-by-step simplification of a given algebraic expression, focusing on techniques such as factorization, exponent manipulation, and algebraic identities. Our goal is to break down the expression:

$\left(\frac{m+\left(m n^2\right)^{\frac{1}{3}}+\left(m^2 n\right)^{\frac{1}{3}}}{m-n}\right) \times\left(1-\frac{n^{\frac{1}{3}}}{m^{\frac{1}{3}}}\right)$

into its simplest form. This exploration will not only provide a solution but also illuminate the underlying mathematical principles, making it easier to tackle similar problems in the future. Understanding how to simplify expressions is crucial in various fields, from calculus to physics, and is an essential tool in any mathematician's arsenal.

To effectively simplify the given expression, we need to dissect it into manageable parts and apply appropriate algebraic techniques. Let's begin by rewriting the terms with fractional exponents in a more familiar radical form. This will make it easier to visualize and manipulate the terms. Fractional exponents can often be a hurdle, so converting them to radicals can provide a clearer path forward.

The given expression is:

$\left(\frac{m+\left(m n^2\right)^{\frac{1}{3}}+\left(m^2 n\right)^{\frac{1}{3}}}{m-n}\right) \times\left(1-\frac{n^{\frac{1}{3}}}{m^{\frac{1}{3}}}\right)$

First, let's rewrite the terms $\left(m n^2\right)^{\frac{1}{3}}$ and $\left(m^2 n\right)^{\frac{1}{3}}$ using radicals:

$\left(m n^2\right)^{\frac{1}{3}} = \sqrt[3]{m n^2}$

$\left(m^2 n\right)^{\frac{1}{3}} = \sqrt[3]{m^2 n}$

Now, let's rewrite $n^{\frac{1}{3}}$ and $m^{\frac{1}{3}}$ as cube roots:

$n^{\frac{1}{3}} = \sqrt[3]{n}$

$m^{\frac{1}{3}} = \sqrt[3]{m}$

Substituting these back into the original expression, we get:

$\left(\frac{m+\sqrt[3]{m n^2}+\sqrt[3]{m^2 n}}{m-n}\right) \times\left(1-\frac{\sqrt[3]{n}}{\sqrt[3]{m}}\right)$

This form allows us to see the structure more clearly and identify potential algebraic manipulations. The next step involves looking for patterns or identities that can help us simplify the numerator and the fraction on the right. The goal is to transform the expression into a form where cancellations or simplifications become apparent. The process of simplifying radical expressions often involves recognizing underlying patterns and applying algebraic identities strategically.

In this section, we will delve into the application of algebraic identities to simplify the expression. The key to simplifying this expression lies in recognizing a pattern that resembles the sum of cubes factorization. The sum of cubes identity is:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Our expression contains terms that can be related to this identity. Notice that the numerator of the first fraction has terms that look like the $a^2 + ab + b^2$ part of the identity. Let's rewrite the expression to make this connection clearer.

The expression we have is:

$\left(\frac{m+\sqrt[3]{m n^2}+\sqrt[3]{m^2 n}}{m-n}\right) \times\left(1-\frac{\sqrt[3]{n}}{\sqrt[3]{m}}\right)$

Let's express $m$ and $n$ as cubes of their cube roots:

$m = (\sqrt[3]{m})^3$

$n = (\sqrt[3]{n})^3$

Substitute these back into the denominator of the first fraction:

$m - n = (\sqrt[3]{m})^3 - (\sqrt[3]{n})^3$

Now, we can apply the difference of cubes factorization to the denominator:

$(\sqrt[3]{m})^3 - (\sqrt[3]{n})^3 = (\sqrt[3]{m} - \sqrt[3]{n})((\sqrt[3]{m})^2 + \sqrt[3]{m}\sqrt[3]{n} + (\sqrt[3]{n})^2)$

Simplifying the terms inside the second parenthesis:

$(\sqrt[3]{m} - \sqrt[3]{n})(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})$

Now, let's rewrite the numerator of the first fraction in a similar form:

$m+\sqrt[3]{m n^2}+\sqrt[3]{m^2 n} = (\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}$

We can also rewrite the second term in the original expression:

$1-\frac{\sqrt[3]{n}}{\sqrt[3]{m}} = \frac{\sqrt[3]{m} - \sqrt[3]{n}}{\sqrt[3]{m}}$

Now, the entire expression becomes:

$\left(\frac{(\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m} - \sqrt[3]{n})(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})}\right) \times \left(\frac{\sqrt[3]{m} - \sqrt[3]{n}}{\sqrt[3]{m}}\right)$

Notice that the term $(\sqrt[3]{m} - \sqrt[3]{n})$ appears in both the denominator of the first fraction and the numerator of the second fraction. This allows us to cancel these terms out, which is a crucial step in simplifying the expression.

Continuing from our previous step, we have the expression:

$\left(\frac{(\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m} - \sqrt[3]{n})(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})}\right) \times \left(\frac{\sqrt[3]{m} - \sqrt[3]{n}}{\sqrt[3]{m}}\right)$

We can cancel out the $(\sqrt[3]{m} - \sqrt[3]{n})$ terms from the numerator and denominator:

$\frac{(\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})} \times \frac{1}{\sqrt[3]{m}}$

Now, let's focus on the numerator of the first fraction:

$(\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n} = m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}$

And the denominator of the first fraction:

$\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2}$

We can rewrite the expression as:

$\frac{m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})\sqrt[3]{m}}$

Now, we can rewrite the numerator of the first fraction as:

$m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n} = (\sqrt[3]{m})^3 + \sqrt[3]{m}\sqrt[3]{n^2} + \sqrt[3]{m^2}\sqrt[3]{n}$

$\qquad \qquad \qquad \qquad \qquad= (\sqrt[3]{m})^3 + \sqrt[3]{m}(\sqrt[3]{n})^2 + (\sqrt[3]{m})^2\sqrt[3]{n}$

Notice that the denominator of the first fraction resembles part of the difference of cubes factorization we used earlier. Let's multiply the numerator and denominator of the entire expression by $\sqrt[3]{m}$:

$\frac{m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})\sqrt[3]{m}} \times \frac{\sqrt[3]{m}}{\sqrt[3]{m}} = \frac{\sqrt[3]{m}(m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n})}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})m}$

This multiplication might seem counterintuitive, but it helps us to further simplify the expression by creating a common factor.

After multiplying the numerator and the denominator by $\sqrt[3]{m}$, we have:

$\frac{\sqrt[3]{m}(m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n})}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})m}$

Let's expand the numerator:

$\sqrt[3]{m}(m + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}) = m\sqrt[3]{m} + \sqrt[3]{m^2 n^2 m} + \sqrt[3]{m^3 n} = m\sqrt[3]{m} + m\sqrt[3]{n^2} + m\sqrt[3]{n}$

Now, let's rewrite the denominator:

$(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})m$

So, the expression becomes:

$\frac{m\sqrt[3]{m} + m\sqrt[3]{n^2} + m\sqrt[3]{n}}{m(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})}$

We can factor out $m$ from the numerator:

$\frac{m(\sqrt[3]{m} + \sqrt[3]{n^2} + \sqrt[3]{n})}{m(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})}$

Now, we can cancel out the $m$ terms:

$\frac{\sqrt[3]{m} + \sqrt[3]{n^2} + \sqrt[3]{n}}{\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2}}$

Now let $a=\sqrt[3]{m}$ and $b=\sqrt[3]{n}$. The expression becomes:

$\frac{a+b^2+b}{a^2+ab+b^2}$

This expression does not simplify further without additional context or constraints on $m$ and $n$. However, let's revisit an earlier step where we had:

$\frac{(\sqrt[3]{m})^3 + \sqrt[3]{m n^2} + \sqrt[3]{m^2 n}}{(\sqrt[3]{m^2} + \sqrt[3]{mn} + \sqrt[3]{n^2})} \times \frac{1}{\sqrt[3]{m}}$

Substituting $a$ and $b$:

$\frac{a^3 + ab^2 + a^2b}{(a^2 + ab + b^2)} \times \frac{1}{a}$

Factoring out $a$ from the numerator:

$\frac{a(a^2 + b^2 + ab)}{(a^2 + ab + b^2)} \times \frac{1}{a}$

$\frac{a(a^2 + ab + b^2)}{a(a^2 + ab + b^2)}$

Cancelling out the common terms, we get 1.

Therefore, the simplified expression is:

$1$

In conclusion, by meticulously applying algebraic identities, factoring, and cancelling out terms, we have successfully simplified the given expression to 1. This exercise underscores the importance of recognizing patterns, understanding algebraic manipulations, and systematically breaking down complex problems into manageable steps. The ability to simplify expressions is a cornerstone of mathematical proficiency and is invaluable in various scientific and engineering disciplines.