Correct Answer: 1

Solution : Given: $xy = -6$ and $x^3+ y^3= 19$ ($x$ and $y$ are integers)
We know the algebraic identity, $(x+y)^3=x^3 + y^3+3xy(x+y)$.
Substitute the given values in the above formula,
⇒ $(x+y)^3=19+3\times(-6)\times(x+y)$
⇒ $(x+y)^3=19-18\times(x+y)$
Let $(x+y)=u$.
⇒ $u^3+18u–19=0$
⇒ $u^3–u^2+u^2–u+19u–19=0$
⇒ $u^2(u–1)+u(u–1)+19(u–1)=0$
⇒ $(u–1)(u^2+u+19)=0$
⇒ $u-1=0$
⇒ $u=1$
⇒ $\frac{1}{x^{–1}}+\frac{1}{y^{–1}}=x+y=1$
Hence, the correct answer is 1.