Question

The cube of $xy-\frac{1}{(xy{)}^2}=$

• A. $x^3{y}^3-\frac{1}{x^6{y}^6}-3+\frac{3}{xy}$
• B. $x^3{y}^3-\frac{1}{x^6{y}^6}-3+\frac{3}{x^2{y}^2}$
• C. $x^3{y}^3-\frac{1}{x^6{y}^6}-3+\frac{3}{x^3{y}^3}$
• D. $x^3{y}^3-\frac{1}{x^3{y}^3}-3+\frac{3}{x^3{y}^3}$

Answer 1

The correct option is C

$x^3{y}^3-\frac{1}{x^6{y}^6}-3+\frac{3}{x^3{y}^3}$

Cube of $xy-\frac{1}{(xy{)}^2}={(xy-\frac{1}{(xy{)}^2})}^3$

Using identity: $(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$, we get

${(xy-\frac{1}{(xy{)}^2})}^3=(xy{)}^3-{\left(\frac{1}{(xy{)}^2}\right)}^3-3\times (xy{)}^2\times \frac{1}{(xy{)}^2}+3\times xy\times {\left(\frac{1}{(xy{)}^2}\right)}^2$

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

$=x^3{y}^3-\frac{1}{x^6{y}^6}-3+\frac{3}{x^3{y}^3}$