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Chapter 9: Problem 40

Use the Binomial Theorem to expand and simplify the expression. \(\left(3-y^{2}\right)^{3}\)

Short Answer

Expert verified
The expanded form of the expression \(\left(3-y^{2}\right)^{3}\) is \(27 - 27y^2 + 27y^4 - y^6\).

Step by step solution

01

Apply the Binomial Theorem

Write the expression \(\left(3-y^{2}\right)^{3}\) using the binomial theorem. According to the binomial theorem:\(\left(3-y^{2}\right)^{3} = \sum_{k=0}^{3} {3 \choose k} (3)^{3-k} (-y^{2})^k\)
02

Expand the Expression

Now, expand the expression by calculating each term of the summation: \({3 \choose 0} (3)^{3-0} (-y^{2})^0+{3 \choose 1}(3)^{3-1}(-y^{2})^1+{3 \choose 2}(3)^{3-2}(-y^{2})^2+{3 \choose 3}(3)^{3-3}(-y^{2})^3\)
03

Simplify Each Term

Calculate the result for each term one by one: \({3 \choose 0} (3)^{3-0} (-y^{2})^0 = 27\),\({3 \choose 1}(3)^{3-1}(-y^{2})^1 = -27y^2\),\({3 \choose 2}(3)^{3-2}(-y^{2})^2 = 27y^4\), and \({3 \choose 3}(3)^{3-3}(-y^{2})^3 = -y^6\).
04

Write the Final Answer

So, \(\left(3-y^{2}\right)^{3} = 27 - 27y^2 + 27y^4 - y^6\).

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Key Concepts

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

Polynomial Expansion
When dealing with the problem of expanding expressions raised to a power, the Binomial Theorem is a powerful tool. It provides a structured way to express powers of binomials as the sum of terms. Each term consists of a combination of coefficients from the binomial series and the powers of the individual terms.
To apply the Binomial Theorem:
  • Identify the binomial, which in this case is \((3 - y^2)^3\).
  • You will express the binomial as the sum of a series of terms using the formula \((a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k\).
  • Here, \(a = 3\), \(b = -y^2\), and \(n = 3\) as shown in the original expression.
Breaking down the binomial using this theorem helps manage the calculation of complex expressions into simpler, manageable parts.
Combinatorics
Combinatorics is an area of mathematics focused on counting, arrangement, and combination of sets. When expanding binomials, combinatorics plays an essential role. It helps in determining the coefficients for each term in the binomial expansion.
This is achieved by using the binomial coefficient \({n \choose k}\), which is read as "n choose k". It calculates the number of ways to select a subset of \(k\) elements from a set of \(n\) elements regardless of the order of selection.
  • The formula for computing binomial coefficients is \({n \choose k} = \frac{n!}{k!(n-k)!}\).
  • These coefficients appear in front of each term after expansion and adjust each term's weight in the final polynomial.
  • In the exercise, coefficients \({3 \choose 0}\), \({3 \choose 1}\), \({3 \choose 2}\), and \({3 \choose 3}\) were calculated to expand \((3 - y^2)^3\).
Combining arithmetic with combinatorics allows us to find the precise number needed for each term in the expansion.
Algebraic Simplification
Once the polynomial expansion is complete, the next step is algebraic simplification. This involves reducing the expanded expression into its simplest form. By simplifying each term in the expanded polynomial, the solution becomes easier to understand and more mathematically elegant.
To simplify, follow these steps:
  • First, compute the powers of constants and coefficients as well as fundamental operations on powers of variables.
  • This includes transforming terms like \((3)^{3-k} (-y^2)^k\) into complete numbers or simplified powers such as \(27, -27y^2, 27y^4,\) and \(-y^6\).
  • Finally, combine these terms to write down the expanded form succinctly: \(27 - 27y^2 + 27y^4 - y^6\).
Simplification is the cherry on top of algebraic operations, presenting a clean and comprehensible result.

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Most popular questions from this chapter

Writing the Terms of a Geometric Sequence Write the first five terms of the geometric sequence. $$a_{1}=5, r=-\frac{1}{10}$$

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Write an expression for the apparent \(n\) th term of the sequence. (Assume \(n\) begins with \(1 .\)) $$1,3,1,3,1, . . .$$

Find the binomial coefficient. \(\left(\begin{array}{l}12 \\ 0\end{array}\right)\)

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