If $x-y=11$ and $\frac{1}{x}-\frac{1}{y}=\frac{11}{24}$, then what is the value of $x^3-y^3+x^2 y^2 ?$
Answer & explanation
Correct answer: option 1
If x - y = n
then, $x^3 - y^3$ = n3 + 3 × n × xy
$x-y=11$
$\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$,
Then what is the value of $x^3-y^3+x^2 y^2 ?$
$\frac{1}{x}-\frac{1}{y}=\frac{11}{24}$,
\(\frac{y - x}{xy}\) = $\frac{11}{24}$
\(\frac{-11}{xy}\) = $\frac{11}{24}$
xy = -24
Then, $\left(x^3+y^3\right) $ = 43 - 3 × 4 × $\frac{15}{4}$]
$x^3 - y^3$ = 113 + 3 × 11 × -24 = 539
$x^3-y^3+x^2 y^2 $ = 539 + (-24 × -24) = 1115