Question

simplify (x-2/3y)³ (x+2/3y)³

Answer 1

$\textbf{Answer:}$

(x-2/3y)³ - (x+2/3y)³

=[x³-(2/3y)³-3(x)(2/3y) (x-2/3y)

=[x³+2/3y)³+3(x)(2/3y) ( x+2/y)

=[x³-8/27y³-2x²y+4/3xy²)-[x³+8/27y³+2x²y+4/3xy²)

=-16/27y³-4x²y



$\textbf{Hope it helps you!}$

Answer 2

Answer:

simplified form is $\frac{729x^6y^6-972x^4y^4+432x^2y^2-64}{729y^6}$

Step-by-step explanation:

we have to simplify the expression $(x-\frac{2}{3y})^{3}(x+\frac{2}{3y})^{3}$

Since, $(a-b)^{3}=a^{3}+b^{3}-3a^{2}b+3ab^{2}-b^{3}$ and  $(a-b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}+b^{3}$

$(x-\frac{2}{3y})^{3}(x+\frac{2}{3y})^{3}$    .......(1)

$\left(x-\frac{2}{3y}\right)^3=x^3-\frac{2x^2}{y}+\frac{4x}{3y^2}-\frac{8}{27y^3}$

$\left(x+\frac{2}{3y}\right)^3=x^3+\frac{2x^2}{y}+\frac{4x}{3y^2}+\frac{8}{27y^3}$

Now, put these in equation ( 1 )

$=\left(\frac{4x}{3y^2}+x^3-\frac{2x^2}{y}-\frac{8}{27y^3}\right)\left(\frac{2x^2}{y}+\frac{4x}{3y^2}+\frac{8}{27y^3}+x^3\right)$

Simplify,

$=\frac{729x^6y^6-972x^4y^4+432x^2y^2-64}{729y^6}$