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Chapter 8: Problem 71

Use the Binomial Theorem to expand and then simplify the result: \(\left(x^{2}+x+1\right)^{3}\). [ Hint: Write \(x^{2}+x+1\) as \(\left.x^{2}+(x+1)\right]\)

Short Answer

Expert verified
\(\left(x^{2}+x+1\right)^{3}\) expands to \(x^6 + 3x^5 + 6x^4 + 10x^3 + 9x^2 + 3x + 1\) using the binomial theorem

Step by step solution

01

Rewrite The Expression

In order to expand the expression, first rewrite it as \( (x^2 + (x + 1))^3 \) according to the provided hint.
02

Apply The Binomial Theorem

Apply the binomial theorem to the rewritten expression. The binomial theorem is \( (a + b)^n = \Sigma_{k=0}^n \binom{n}{k} a^{n-k} * b^k \). This means that across three terms, you get four terms in the expansion: \[ (x^2 + (x + 1))^3 = \binom{3}{0} (x^2)^3 + \binom{3}{1} (x^2)^2 * (x + 1) + \binom{3}{2} (x^2) (x + 1)^2 + \binom{3}{3} (x + 1)^3 \]
03

Do the Calculations

Calculate the coefficients using the binomial theorem values and simplify the algebraic expressions. This leads to the following results:\[ = 1 * x^6 + 3 * x^4 * (x + 1) + 3 * x^2 * (x + 1)^2 + 1 * (x + 1)^3 \]\[ = x^6 + 3x^5 + 3x^4 + 3x^4 + 6x^3 + 3x^2 + x^3 + 3x^2 + 3x + 1 \]\[ = x^6 + 3x^5 + 6x^4 + 10x^3 + 9x^2 + 3x + 1 \]
04

Final Simplification

Combine like terms in the resulting expression from the previous step. Then the expanded and simplified expression becomes:\[ x^6 + 3x^5 + 6x^4 + 10x^3 + 9x^2 + 3x + 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expansion using the Binomial Theorem
The Binomial Theorem is a powerful tool for expanding expressions of the form \((a + b)^n\). It states that such an expression can be expanded into a sum involving terms of the form \(\binom{n}{k} a^{n-k} b^k\) where the \(\binom{n}{k}\) is a binomial coefficient. When approaching the problem of expanding \((x^2 + (x+1))^3\), we follow these steps.

First, identify the terms: \(a\) is \(x^2\) and \(b\) is \(x + 1\). The exponent \(n\) is 3. Apply the theorem to get:
  • \(\binom{3}{0} (x^2)^3\)
  • \(\binom{3}{1} (x^2)^2 (x + 1)\)
  • \(\binom{3}{2} (x^2) (x + 1)^2\)
  • \(\binom{3}{3} (x + 1)^3\)
This results in an expanded polynomial expression that still needs simplification.
Polynomial Simplification
After obtaining the expanded terms using the Binomial Theorem, the next step is simplifying each term. This involves several algebraic operations:

  • Raise the individual powers and compute coefficients.
  • Multiply terms to handle brackets if present.
  • Combine like terms effectively.
For the given expansion:
  • Calculate each term such as \((x^2)^2 * (x+1)\) and expand further.
  • Combine like terms to simplify the expression to \(x^6 + 3x^5 + 6x^4 + 10x^3 + 9x^2 + 3x + 1\).
The aim of simplification is to condense the polynomial expression into a form with no redundant or unnecessary operations, making calculations easier and cleaner.
Combinatorics in the Binomial Theorem
Combinatorics plays a crucial role in the Binomial Theorem by providing the binomial coefficients \(\binom{n}{k}\). These coefficients are derived from combinatorial principles like permutations and combinations. To calculate \(\binom{3}{k}\) for \(k = 0, 1, 2, 3\), use the formula:
  • \(\binom{3}{0} = 1\)
  • \(\binom{3}{1} = 3\)
  • \(\binom{3}{2} = 3\)
  • \(\binom{3}{3} = 1\)
These coefficients correspond to the weights for each term in the expansion and reflect the number of ways to choose subsets of terms from the binomial expression \((a + b)^n\). The central idea is that combinatorial counts simplify polynomial expansion calculations, making them manageable and intuitive.
Understanding Powers of Polynomials
Powers of polynomials refer to raising a polynomial expression to a given power, as seen in expressions like \((x^2 + x + 1)^3\). When raising a polynomial to a power, each individual term within the polynomial is raised to that power and combined in a specific binomial pattern.

For example, when considering \( (x^2)\) in \((x^2 + (x + 1))^3\), the powers will vary depending on their position in the expansion term set by the Binomial Theorem:
  • \((x^2)^3\) contributes the highest power, \(6\).
  • Mixed terms like \((x^2)^2(x+1)\) contribute middle power terms.
  • \((x+1)^3\) contributes lower power combinations due to shared distribution.
Each term's power and combination reflect the polynomial arithmetic performed during expansion, assisting in achieving the eventual simplified form.

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Most popular questions from this chapter

Follow the outline on the next page to use mathematical induction to prove that $$ \begin{aligned}(a+b)^{n}=\left(\begin{array}{c}n \\\0\end{array}\right) a^{n}+\left(\begin{array}{c}n \\\1\end{array}\right) a^{n-1} b+\left(\begin{array}{c}n \\\2\end{array}\right) a^{n-2} b^{2} \\\\+\cdots+\left(\begin{array}{c}n \\\n-1\end{array}\right) a b^{n-1}+\left(\begin{array}{c}n \\\n\end{array}\right) b^{n}\end{aligned} $$ a. Verify the formula for \(n=1\) b. Replace \(n\) with \(k\) and write the statement that is assumed true. Replace \(n\) with \(k+1\) and write the statement that must be proved. c. Multiply both sides of the statement assumed to be true by \(a+b .\) Add exponents on the left. On the right, distribute \(a\) and \(b,\) respectively. d. Collect like terms on the right. At this point, you should have $$ \begin{aligned}&(a+b)^{k+1}=\left(\begin{array}{l}k \\\0\end{array}\right) a^{k+1}+\left[\left(\begin{array}{l}k \\\0\end{array}\right)+\left(\begin{array}{l}k \\\1\end{array}\right)\right] a^{k} b\\\&\begin{array}{l}+\left[\left(\begin{array}{c}k \\\1\end{array}\right)+\left(\begin{array}{c}k \\\2\end{array}\right)\right] a^{k-1} b^{2}+\left[\left(\begin{array}{c}k \\\2\end{array}\right)+\left(\begin{array}{c}k \\\3\end{array}\right)\right] a^{k-2} b^{3} \\\\+\cdots+\left[\left(\begin{array}{c}k \\\k-1\end{array}\right)+\left(\begin{array}{c}k \\\k\end{array}\right)\right] a b^{k}+\left(\begin{array}{c}k \\\k\end{array}\right) b^{k+1} \end{array}\end{aligned} $$ e. Use the result of Exercise 74 to add the binomial sums in brackets. For example, because \(\left(\begin{array}{l}n \\\ r\end{array}\right)+\left(\begin{array}{c}n \\\ r+1\end{array}\right)$$=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\) then \(\left(\begin{array}{l}k \\ 0\end{array}\right)+\left(\begin{array}{l}k \\\ 1\end{array}\right)=\left(\begin{array}{c}k+1 \\\1\end{array}\right)\) and\(\left(\begin{array}{l}k \\ 1\end{array}\right)+\left(\begin{array}{l}k \\\2\end{array}\right)=\left(\begin{array}{c}k+1 \\ 2\end{array}\right)\) f. Because \(\left(\begin{array}{l}k \\\ 0\end{array}\right)=\left(\begin{array}{c}k+1 \\ 0\end{array}\right) \quad\) (why?) and \(\left(\begin{array}{l}k \\ k\end{array}\right)=\) \(\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)\) (why?), substitute these results and the results from part (e) into the equation in part (d). This should give the statement that we were required to prove in the second step of the mathematical induction process.

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (y-3)^{4} $$

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (c+2)^{5} $$

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (x+4)^{3} $$

Describe the pattern on the exponents on \(a\) in the expansion of \((a+b)^{n}\).
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