Question

State True or False.

${[x-\frac{2}{3}y]}^3$$=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$

• A. True
• B. False

Answer 1

The correct option is A True

Using, $(a+b{)}^3=a^3+b^3+3ab(a+b)$
$(x-\frac{2}{3}y{)}^3=(x{)}^3+(-\frac{2}{3}y{)}^3+3(x\left)\right(-\frac{2}{3}y\left)\right(x-\frac{2}{3}y)$
$=x^3-\frac{8}{27}{y}^3-2xy(x-\frac{2}{3}y)$
$=x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$