Question

Subtract the following:

$\frac{1}{4}{x}^3+xy+x^2+\frac{1}{6}{y}^3$ from $x^3+\frac{2}{3}xy+x^2-\frac{7}{6}-6y^3$

Answer: $\frac{3}{4}{x}^3-\frac{1}{3}xy-\frac{4}{3}{y}^3$

• A. True
• B. False

Answer 1

The correct option is B False

$x^3+\frac{2}{3}xy+x^2-\frac{7}{6}-6y^3$

$-(\frac{1}{4}{x}^3+xy+x^2+\frac{1}{6}{y}^3)$

$=x^3-\frac{1}{4}{x}^3+\frac{2}{3}xy-xy+x^2-x^2-\frac{7}{6}-6y^3-\frac{1}{6}{y}^3$

$=\frac{3}{4}{x}^3-\frac{1}{3}xy-\frac{7}{6}-\frac{37}{6}{y}^3$

This doesn't match with given answer.

So, FALSE