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Chapter 7: Problem 44

Use Pascal's triangle to simplify the indicated expression. $$ (3-\sqrt{2})^{6} $$

Short Answer

Expert verified
Using Pascal's Triangle and the binomial theorem, we find that \((3-\sqrt{2})^6 = 3^6 - 6\cdot 3^5\sqrt{2} + 30\cdot 3^4 - 40\cdot 3^3\sqrt{2} + 60\cdot3^2 - 48\cdot 3\sqrt{2} + 8\).

Step by step solution

01

Write down the 6th row of Pascal's Triangle

The first few rows of Pascal's Triangle look like this: ``` 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 ``` In our case, we need the 6th row (excluding the 0th one), which is: \(1, 6, 15, 20, 15, 6, 1\).
02

Use the binomial theorem to expand the expression

The binomial theorem allows us to expand \((a+b)^n\) as follows: \((a + b)^n = C_{n,0}\cdot a^{n} b^{0}+ C_{n,1}\cdot a^{n-1} b^1 + C_{n,2}\cdot a^{n-2} b^{2} + \cdots + C_{n,n} \cdot a^{0} b^{n}\) where \(C_{n,k}\) is the binomial coefficient, the numbers in Pascal's Triangle. Applying the binomial theorem to our problem with \(a=3, b=-\sqrt{2}\), and \(n=6\), we get: \((3-\sqrt{2})^6 = \sum_{k=0}^{6} C_{6,k} \cdot 3^{6-k} (-\sqrt{2})^k\)
03

Calculate each term in the expansion

Let's compute every term in the expansion: 1. \(C_{6,0} \cdot 3^{6} (-\sqrt{2})^0 = 1\cdot 3^6\cdot 1 = 3^6\) 2. \(C_{6,1} \cdot 3^{5} (-\sqrt{2})^1 = 6\cdot 3^5\cdot(-\sqrt{2}) =-6\cdot3^5\sqrt{2}\) 3. \(C_{6,2} \cdot 3^{4} (-\sqrt{2})^2 = 15\cdot 3^4\cdot 2= 30\cdot 3^4\) 4. \(C_{6,3} \cdot 3^{3} (-\sqrt{2})^3 = 20\cdot 3^3\cdot(-2\sqrt{2}) = -40\cdot 3^3\sqrt{2}\) 5. \(C_{6,4} \cdot 3^{2} (-\sqrt{2})^4 = 15\cdot 3^2\cdot 4 = 60\cdot3^2\) 6. \(C_{6,5} \cdot 3^{1} (-\sqrt{2})^5 = 6\cdot 3^1\cdot(-8\sqrt{2}) = -48\cdot 3\sqrt{2}\) 7. \(C_{6,6} \cdot 3^{0} (-\sqrt{2})^6 = 1\cdot 1\cdot 8 = 8\)
04

Add all terms together

Finally, let's add all terms together: \((3-\sqrt{2})^6 = 3^6 - 6\cdot 3^5\sqrt{2} + 30\cdot 3^4 - 40\cdot 3^3\sqrt{2} + 60\cdot3^2 - 48\cdot 3\sqrt{2} + 8\) This is the simplified expression of \((3-\sqrt{2})^6\), given by Pascal's Triangle and the binomial theorem.

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