Question

State True or False.

$(x-2y)(\frac{1}{3}{x}^2-\frac{1}{2}xy+\frac{1}{6}{y}^2)$$=\frac{1}{3}{x}^3-\frac{7}{6}{x}^{2}y+\frac{7}{6}x{y}^2-\frac{1}{3}{y}^3$

• A. True
• B. False

Answer 1

The correct option is A True

Multiplying term by term and then adding,

$(x-2y)(\frac{1}{3}{x}^2-\frac{1}{2}xy+\frac{1}{6}{y}^2)=x(\frac{1}{3}{x}^2-\frac{1}{2}xy+\frac{1}{6}{y}^2)-2y(\frac{1}{3}{x}^2-\frac{1}{2}xy+\frac{1}{6}{y}^2)=\frac{1}{3}{x}^3-\frac{1}{2}{x}^{2}y+\frac{1}{6}x{y}^2-\frac{2}{3}{x}^{2}y+x{y}^2-\frac{1}{3}{y}^3=\frac{1}{3}{x}^3-\frac{7}{6}{x}^{2}y+\frac{7}{6}x{y}^2-\frac{1}{3}{y}^3$