Question

1. expand ((5x - 4y)^3) by applying the binomial theorem.

(\bigcirc) (125x^3 + 200x^2y + 160xy^2 + 64y^3)
(\bigcirc) (125x^3 - 200x^2y + 160xy^2 - 64y^3)
(\bigcirc) (125x^3 + 300x^2y - 240xy^2 - 64y^3)
(\bigcirc) (125x^3 - 300x^2y + 240xy^2 - 64y^3)

Explanation:

Step 1: Recall the binomial theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\). For \((5x-4y)^{3}\), we have \(a = 5x\), \(b=-4y\) and \(n = 3\).

Step 2: Calculate each term

• When \(k = 0\):

\(\binom{3}{0}(5x)^{3}(-4y)^{0}=\frac{3!}{0!(3 - 0)!}\times125x^{3}\times1 = 1\times125x^{3}=125x^{3}\)

• When \(k = 1\):

\(\binom{3}{1}(5x)^{2}(-4y)^{1}=\frac{3!}{1!(3 - 1)!}\times25x^{2}\times(-4y)=3\times25x^{2}\times(-4y)=- 300x^{2}y\)

• When \(k = 2\):

\(\binom{3}{2}(5x)^{1}(-4y)^{2}=\frac{3!}{2!(3 - 2)!}\times5x\times16y^{2}=3\times5x\times16y^{2}=240xy^{2}\)

• When \(k = 3\):

\(\binom{3}{3}(5x)^{0}(-4y)^{3}=\frac{3!}{3!(3 - 3)!}\times1\times(-64y^{3})=1\times(-64y^{3})=-64y^{3}\)

Step 3: Combine the terms

Adding up all the terms: \(125x^{3}-300x^{2}y + 240xy^{2}-64y^{3}\)

Answer:

D. \( 125x^{3}-300x^{2}y + 240xy^{2}-64y^{3} \)