Problem 84 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the Binomial Theorem to expand and simplify the expression. \(\left(u^{3 / 5}+v^{1 / 5}\right)^{5}\)

Step-by-Step Solution

Answer

The simplified expression of \(\left(u^{3 / 5}+v^{1 / 5}\right)^{5}\) using the Binomial Theorem is \(u^3 + 5u^{12/5}v^{1/5} + 10u^{6/5}v^{2/5} + 10u^{3/5}v^{3/5} + 5u^{3/5}v^{4/5} + v\)

Step 1: Recall the Binomial Theorem

The Binomial theorem states that \((x+y)^n= \sum_{k=0}^{n} {n \choose k} x^{n-k}y^k\)

Step 2: Apply the Binomial Theorem

Applying this to the given expression, we get \(\left(u^{3 / 5}+v^{1 / 5}\right)^{5}= \sum_{k=0}^{5} {5 \choose k} (u^{3 / 5})^{5-k}(v^{1 / 5})^k\)

Step 3: Expand using Binomial coefficients

The expansion, using the binomial coefficients from Pascal's triangle, becomes \(u^3 + 5*u^{12/5}*v^{1/5} + 10*u^{6/5}*v^{2/5} + 10*u^{3/5}*v^{3/5} + 5*u^{3/5}*v^{4/5} + v\)

Step 4: Simplify the Expression

For the final expression, simplify wherever possible to obtain : \(u^3 + 5u^{12/5}v^{1/5} + 10u^{6/5}v^{2/5} + 10u^{3/5}v^{3/5} + 5u^{3/5}v^{4/5} + v\)

Key Concepts

Binomial Coefficients

When we talk about the Binomial Theorem, binomial coefficients are at the heart of understanding how to expand expressions like \( (x+y)^n \). These coefficients are the numbers that multiply the variables in the expansion. They can be computed using the formula \( {n \choose k} \), which is read as 'n choose k'. What this calculates is the number of ways to choose k elements out of a set of n elements, without considering the order.

These coefficients have several properties that make them special: they are symmetrical for any given n, and they form the well-known pattern called Pascal's triangle. When expanding \( (u^{3 / 5}+v^{1 / 5})^{5} \), we'd use the coefficients corresponding to n = 5 in Pascal's triangle: 1, 5, 10, 10, 5, and 1, as seen in the solved exercise.

Pascal's Triangle

Pascal's triangle is a simple yet powerful tool in combinatorics, named after its pioneer Blaise Pascal. This triangular array starts with a '1' at the top and with each new row, every number is the sum of the two numbers directly above it in the previous row. It can be used to quickly find the binomial coefficients needed for polynomial expansion.

For instance, if we want to determine the coefficients for the expansion of \( (x+y)^n \), we simply look at the nth row of Pascal's triangle. In our exercise, since the exponent is 5, we take the sixth row (remember, the top of the triangle is row 0), which gives us the coefficients 1, 5, 10, 10, 5, and 1. This triangle is not only useful for binomial expansion but also has applications in probability and other areas of mathematics.

Polynomial Expansion

Polynomial expansion is the process of breaking down a raised binomial expression into a series of terms consisting of coefficients multiplied by powers of the variables in the binomial. It's particularly streamlined by the Binomial Theorem, which allows us to swiftly expand any binomial expression without having to manually multiply the binomial by itself n times.

In the given exercise example, \( \left(u^{3 / 5}+v^{1 / 5}\right)^{5} \) is expanded into \( u^3 + 5u^{12/5}v^{1/5} + 10u^{6/5}v^{2/5} + 10u^{3/5}v^{3/5} + 5u^{3/5}v^{4/5} + v \). Each term in this expansion represents a possible combination of the terms \( u^{3 / 5} \) and \( v^{1 / 5} \) multiplied together, following the order given by the Binomial Theorem and multiplied by the corresponding coefficients from Pascal's triangle.

Exponent Simplification

After expansion, it's not unusual to encounter terms with exponents that can be simplified to make the expression easier to understand or to use in further calculations. Simplification of exponents involves finding equivalent expressions where the powers are reduced to lower terms or whole numbers if possible. Often, this means applying the basic rules of exponents, such as \( (x^m)^n = x^{mn} \) and \( x^m \cdot x^n = x^{m+n} \).

In our textbook exercise, the process of simplifying powers of \( u^{3 / 5} \) and \( v^{1 / 5} \) after they have been raised to additional powers is crucial. As an example, taking \( (u^{3 / 5})^2 \) results in \( u^{6/5} \) because we multiply the exponents (3/5 and 2), demonstrating exponent simplification at work.