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Chapter 11: Problem 11

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

Short Answer

Expert verified
The expanded form of \((3x + y)^3\) is \(27x^{3}\) + \(27x^{2}y\) + \(9xy^{2}\) + \(y^{3}\).

Step by step solution

01

Apply the Binomial Theorem

The Binomial Theorem is expressed as \((a + b)^n = \sum_{k=0}^{n} C_{n}^{k} a^{n-k} b^{k}\). Apply this formula to our given expression \((3x + y)^3\). Replace \(a\) with \(3x\), \(b\) with \(y\), and \(n\) with 3. This results in \((3x + y)^3 = \sum_{k=0}^{3} C_{3}^{k} (3x)^{3-k} y^{k}\).
02

Calculate each term in the expansion

Now, calculate the four terms in the expression.First term: \(C_{3}^{0} (3x)^{3-0} y^{0}\) = 1 * \(27x^{3}\) * 1 = \(27x^{3}\).Second term: \(C_{3}^{1} (3x)^{3-1} y^{1}\) = 3 * \(9x^{2}\) * y = \(27x^{2}y\).Third term: \(C_{3}^{2} (3x)^{3-2} y^{2}\) = 3 * \(3x\) * \(y^{2}\) = \(9xy^{2}\).Last term: \(C_{3}^{3} (3x)^{3-3} y^{3}\) = 1 * x^0 * \(y^{3}\) = \(y^{3}\).
03

Sum the calculated terms

Add together the calculated terms: \(27x^{3}\) + \(27x^{2}y\) + \(9xy^{2}\) + \(y^{3}\), which is the expanded form of the given expression.

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