Problem 28 - Contemporary Precalculus

Question

Expand and (where possible) simplify the expression. $$\left(3 u-v^{3}\right)^{6}$$

Step-by-Step Solution

Answer

Question: Expand and simplify the expression \((3u-v^3)^6\). Answer: The expanded and simplified expression for \((3u-v^3)^6\) is: $$729u^6 - 1458u^5v^3 + 1215u^4v^6 - 540u^3v^9 + 135u^2v^{12} - 18uv^{15} + v^{18}$$

Step 1: Write down the binomial theorem formula

The binomial theorem formula is: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ For our expression, we have \(a = 3u\), \(b = -v^3\), and \(n = 6\). Now we will apply the binomial theorem formula to expand our expression.

Step 2: Expand the expression using the binomial theorem

Using the binomial theorem, we have: $$(3u-v^3)^6 = \sum_{k=0}^6 \binom{6}{k} (3u)^{6-k} (-v^3)^k$$ Now we will expand the terms in the summation.

Step 3: Calculate the terms in the summation

We will calculate each term separately and then combine them afterward: For \(k = 0\): \(\binom{6}{0}(3u)^6(-v^3)^0 = 1 \cdot (729u^6) \cdot 1 = 729u^6\) For \(k = 1\): \(\binom{6}{1}(3u)^5(-v^3)^1 = 6 \cdot (243u^5) \cdot (-v^3) = -1458u^5v^3\) For \(k = 2\): \(\binom{6}{2}(3u)^4(-v^3)^2 = 15 \cdot (81u^4) \cdot (v^6) = 1215u^4v^6\) For \(k = 3\): \(\binom{6}{3}(3u)^3(-v^3)^3 = 20 \cdot (27u^3) \cdot (-v^9) = -540u^3v^9\) For \(k = 4\): \(\binom{6}{4}(3u)^2(-v^3)^4 = 15 \cdot (9u^2) \cdot (v^{12}) = 135u^2v^{12}\) For \(k = 5\): \(\binom{6}{5}(3u)(-v^3)^5 = 6 \cdot (3u) \cdot (-v^{15}) = -18uv^{15}\) For \(k = 6\): \(\binom{6}{6}(-v^3)^6 = 1 \cdot (v^{18}) = v^{18}\)

Step 4: Combine the terms

Now we will combine all the terms we calculated in Step 3: $$(3u-v^3)^6 = 729u^6 - 1458u^5v^3 + 1215u^4v^6 - 540u^3v^9 + 135u^2v^{12} - 18uv^{15} + v^{18}$$ The final expanded and simplified expression is: $$729u^6 - 1458u^5v^3 + 1215u^4v^6 - 540u^3v^9 + 135u^2v^{12} - 18uv^{15} + v^{18}$$

Key Concepts

Polynomial Expansion

The process of polynomial expansion involves expressing a power of a binomial, such as \((3u-v^3)^6\), as a sum of terms. Each term consists of products of the binomial's components raised to various powers. This expansion leverages the binomial theorem, which provides a formula to expand expressions of the form \((a+b)^n\). The theorem states:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

Here, \(a = 3u\) and \(b = -v^3\). By applying the formula, we systematically calculate each term in the expansion. This step involves:

• Coefficients, which are determined by the binomial coefficient \(\binom{n}{k}\).
• Powers of \(a\) and \(b\), which come from \(a^{n-k}\) and \(b^k\), respectively.

Combinatorics

Combinatorics plays a crucial role in polynomial expansion via the calculation of binomial coefficients. These coefficients are represented by \(\binom{n}{k}\) and are pivotal in determining the multiplicative factors of each term in the expansion. The binomial coefficient \(\binom{n}{k}\) is defined as:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

This formula calculates the number of ways to choose \(k\) elements from a set of \(n\) elements without regard to order. In our polynomial expansion, the coefficients such as 1, 6, 15, 20, are derived from this formula and they determine how each term in the expansion contributes to the final expression.

Algebraic Expression Simplification

Simplifying an algebraic expression means combining like terms and simplifying coefficients to make the expression more concise. In the context of the binomial expansion, after calculating each term, the next step is to combine them into a single expression. For example:

• Calculate each term using \(\binom{n}{k}(3u)^{6-k}(-v^3)^k\).
• Simplify by performing arithmetic operations like multiplication on coefficients and managing exponents.

Finally, we gather all these terms to present the expanded polynomial in its simplest form. The end expression, \(729u^6 - 1458u^5v^3 + 1215u^4v^6 - 540u^3v^9 + 135u^2v^{12} - 18uv^{15} + v^{18}\), represents this simplification, showing the polynomial in its most simplified expanded state.