Use the binomial theorem to expand the expression.
-$\left(\frac{x}{3}+4 y\right)^{3}$
A) $\frac{1}{27} x^{6}+\frac{8}{9} x^{3} y^{3}+64 y^{6}$
B) $\frac{1}{9} x^{3}+\frac{4}{3} x^{2} y+16 x y^{2}+64 y^{3}$
C) $\frac{1}{27} x^{3}+\frac{4}{3} x^{2} y+16 x y^{2}+64 y^{3}$
D) $\frac{1}{9} x^{3}+\frac{4}{3} x^{2} y+64 y^{3}$

Question

Quizplus · Accepted Answer

The binomial theorem states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. For $\left(\frac{x}{3}+4 y\right)^{3}$, expand using the theorem: $(\frac{x}{3})^3 + 3(\frac{x}{3})^2(4y) + 3(\frac{x}{3})(4y)^2 + (4y)^3$, which simplifies to $\frac{1}{27} x^{3}+\frac{4}{3} x^{2} y+16 x y^{2}+64 y^{3}$.

Use the Binomial Theorem to Expand the Expression (x3+4y)3\left(\frac{x}{3}+4 y\right)^{3}(3x​+4y)3 A) 127x6+89x3y3+64y6\frac{1}{27} x^{6}+\frac{8}{9} x^{3} y^{3}+64 y^{6}271​x6+98​x3y3+64y6

Use the Binomial Theorem to Expand the Expression $\left(\frac{x}{3}+4 y\right)^{3}$ A) $\frac{1}{27} x^{6}+\frac{8}{9} x^{3} y^{3}+64 y^{6}$