91Ó°ÊÓ

The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) is the binomial coefficient \(\frac{n!}{k!(n-k)!}\). For this exercise, \(a = \frac{x}{3}\), \(b = -\frac{y}{2}\), and \(n = 4\).

Calculate the binomial coefficients for \(n = 4\): - \(\binom{4}{0} = 1\)- \(\binom{4}{1} = 4\)- \(\binom{4}{2} = 6\)- \(\binom{4}{3} = 4\)- \(\binom{4}{4} = 1\)

Using the formula, expand each term as follows:- \(k = 0\): \(\binom{4}{0} \left(\frac{x}{3}\right)^{4} \left(-\frac{y}{2}\right)^{0} = 1 \cdot \frac{x^4}{81} \cdot 1 = \frac{x^4}{81}\)- \(k = 1\): \(\binom{4}{1} \left(\frac{x}{3}\right)^{3} \left(-\frac{y}{2}\right)^{1} = 4 \cdot \frac{x^3}{27} \cdot \left(-\frac{y}{2}\right) = -\frac{4x^3y}{54} = -\frac{2x^3y}{27}\)- \(k = 2\): \(\binom{4}{2} \left(\frac{x}{3}\right)^{2} \left(-\frac{y}{2}\right)^{2} = 6 \cdot \frac{x^2}{9} \cdot \frac{y^2}{4} = \frac{6x^2y^2}{36} = \frac{x^2y^2}{6}\)- \(k = 3\): \(\binom{4}{3} \left(\frac{x}{3}\right)^{1} \left(-\frac{y}{2}\right)^{3} = 4 \cdot \frac{x}{3} \cdot -\frac{y^3}{8} = -\frac{4xy^3}{24} = -\frac{xy^3}{6}\)- \(k = 4\): \(\binom{4}{4} \left(\frac{x}{3}\right)^{0} \left(-\frac{y}{2}\right)^{4} = 1 \cdot 1 \cdot \frac{y^4}{16} = \frac{y^4}{16}\)

Add all the terms calculated in Step 3 to write the final expanded expression:\[\left(\frac{x}{3} - \frac{y}{2}\right)^4 = \frac{x^4}{81} - \frac{2x^3y}{27} + \frac{x^2y^2}{6} - \frac{xy^3}{6} + \frac{y^4}{16}\]