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Chapter 0: Problem 54

Find each product. $$(3 x+4)^{3}$$

Short Answer

Expert verified
\((3x+4)^3\) expands to \(27x^3 + 108x^2 + 288x + 64\)

Step by step solution

01

Identifying the Form

Identify that we have a binomial \((a+b)\) raised to a power \(n\). In our case, \(a = 3x\), \(b = 4\), and \(n = 3\).
02

Applying the Binomial Theorem

Apply the binomial theorem. The binomial theorem follows the pattern \((a+b)^n = a^n + (n choose 1) \cdot a^{n-1}b + (n choose 2) \cdot a^{n-2}b^2 +.....+ b^n\).
03

Substituting the Values

Substitute \(a = 3x\), \(b = 4\), and \(n = 3\) into the theorem and we have \((3x+4)^3 = (3x)^3 + (3 choose 1) \cdot (3x)^2\cdot4 + (3 choose 2) \cdot (3x)\cdot4^2 + 4^3\).
04

Simplifying the Expression

Simplify this to get the final answer. \( (3x)^3 = 27x^3\), \((3 choose 1) \cdot (3x)^2\cdot4 = 108x^2\), \((3 choose 2) \cdot (3x)\cdot4^2 = 288x\), and \(4^3 = 64\). When we add these numbers together we get \(27x^3 + 108x^2 + 288x + 64\).

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