If $x-y=11$ and $\frac{1}{x}-\frac{1}{y}=\frac{11}{24}$, then what is the value of $x^3-y^3+x^2 y^2 ?$

1115

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