If x + y = 5 and $\frac{1}{x}+\frac{1}{y} = \frac{20}{9}$, then the value of $(x^3 + y^3)$ will be :

$\frac{365}{4}$

We know that,

If x + y  = n

then, $x^3 + y^3$ = n³ - 3 × n × xy

If x + y = 5

$\frac{1}{x}+\frac{1}{y} = \frac{20}{9}$,

then the value of $(x^3 + y^3)$= ?

Solving $\frac{1}{x}+\frac{1}{y} = \frac{20}{9}$

$\frac{y + x}{xy} = \frac{20}{9}$,

= $\frac{5}{xy} = \frac{20}{9}$,

xy = \(\frac{9}{4}\)

then, $x^3 + y^3$ = 5³ - 3 × 5 × \(\frac{9}{4}\) = 125 - \(\frac{135}{4}\) = $\frac{365}{4}$