If x+y=4 and 1/x+1/y=16/15, then what is the value of ≤ft(x^3+y^3) ?

If x + y  = n

then, $x^3 + y^3$ = n³ - 3 × n × xy

$x+y=4$

$\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$,

Then what is the value of $\left(x^3+y^3\right) ?$

$\frac{1}{x}+\frac{1}{y}=\frac{16}{15}$,

\(\frac{x + y}{xy}\) = $\frac{16}{15}$

\(\frac{4}{xy}\) = $\frac{16}{15}$

xy = $\frac{15}{4}$

Then, $\left(x^3+y^3\right) $ = 4³ - 3 × 4 × $\frac{15}{4}$]

$\left(x^3+y^3\right) $ = 64 - 45 = 19