Sure! Let's expand the expression [tex]\(\left( x - \frac{2}{3} y \right)^3\)[/tex] step by step.
1. Write down the expression: We start with the expression [tex]\(\left( x - \frac{2}{3} y \right)^3\)[/tex].
2. Apply the binomial theorem: The binomial theorem states that [tex]\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)[/tex], where [tex]\(\binom{n}{k}\)[/tex] is the binomial coefficient. For our expression, [tex]\(a = x\)[/tex], [tex]\(b = -\frac{2}{3} y\)[/tex], and [tex]\(n = 3\)[/tex].
3. Expand using the binomial theorem: We need to expand [tex]\(\left(x - \frac{2}{3} y\right)^3\)[/tex].
[tex]\[
\left(x - \frac{2}{3} y\right)^3 = \sum_{k=0}^{3} \binom{3}{k} x^{3-k} \left(-\frac{2}{3}y\right)^k
\][/tex]
4. Calculate each term:
- For [tex]\(k=0\)[/tex]:
[tex]\[
\binom{3}{0} x^3 \left(-\frac{2}{3} y\right)^0 = 1 \cdot x^3 \cdot 1 = x^3
\][/tex]
- For [tex]\(k=1\)[/tex]:
[tex]\[
\binom{3}{1} x^2 \left(-\frac{2}{3} y\right) = 3 \cdot x^2 \cdot \left(-\frac{2}{3} y\right) = -2 x^2 y
\][/tex]
- For [tex]\(k=2\)[/tex]:
[tex]\[
\binom{3}{2} x \left(-\frac{2}{3} y\right)^2 = 3 \cdot x \cdot \left(\frac{4}{9} y^2\right) = \frac{4}{3} x y^2
\][/tex]
- For [tex]\(k=3\)[/tex]:
[tex]\[
\binom{3}{3} \left(-\frac{2}{3} y\right)^3 = 1 \cdot \left(-\frac{8}{27} y^3\right) = -\frac{8}{27} y^3
\][/tex]
5. Combine all terms:
[tex]\[
\left( x - \frac{2}{3} y \right)^3 = x^3 - 2 x^2 y + \frac{4}{3} x y^2 - \frac{8}{27} y^3
\][/tex]
So, the expanded form of [tex]\(\left( x - \frac{2}{3} y \right)^3\)[/tex] is:
[tex]\[
x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3
\][/tex]
And that's the detailed step-by-step expansion of the given expression.
