Problem 55 - College Algebra Essentials

Question

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

Step-by-Step Solution

Answer

The product of the expression \((x-3)^{3}\) is \(x^{3} - 9x^{2} + 27x - 27\)

Step 1: Expand the expression

Use the binomial theorem to expand the expression. So, \((x-3)^{3} = \binom{3}{0}x^{3}(-3)^{0} - \binom{3}{1}x^{2}(-3)^{1} + \binom{3}{2}x^{1}(-3)^{2} - \binom{3}{3}x^0(-3)^3 .\

Step 2: Simplify coefficients

Now, simplify the coefficients. The binomial coefficients are \(\binom{3}{0} = 1\), \(\binom{3}{1} = 3\), \(\binom{3}{2} = 3\), and \(\binom{3}{3} = 1\). Now, find the power of -3. \((-3)^0 = 1\), \((-3)^1 = -3\), \((-3)^2 = 9\), and \((-3)^3 = -27\). Substitute these into the equation above.

Step 3: Substitute and Simplify

Substitute the coefficients and numbers back into the expression. This yields \(x^{3} - 3x^{2}*3 + 3x*9 - 27\). The expression simplifies to \(x^{3} - 9x^{2} + 27x - 27\).