Question

If $x+y=a$ and $xy=b$, then the value of $\frac{1}{x^3}+\frac{1}{y^3}$ is:

• A. $a^3-3ab$
• B. $\frac{a^3+3ab}{b^3}$
• C. $\frac{a^3-3ab}{b^3}$
• D. $a^3+3ab$

Answer 1

The correct option is C $\frac{a^3-3ab}{b^3}$

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

$=\frac{[x^3+y^3+3xy(x+y\left)\right]-3xy(x+y)}{x^3{y}^3}$ ...(Add and subtract $3xy(x+y)$)

$=\frac{(x+y{)}^3-3xy(x+y)}{(xy{)}^3}$ ...$(\because (a+b{)}^3=a^3+b^3+3ab(a+b))$

$=\frac{a^3-(3b\times a)}{b^3}$ ...(Given: $x+y=a,xy=b$)

$=\frac{a^3-3ab}{b^3}$