If $2 x^2-7 x+5=0$, then what is the value of $x^3+\frac{125}{8 x^3}$ ?

$16 \frac{5}{8}$

If x + y  = n

then, $x^3 + y^3$ = n³ - 3 × n × xy

If $2 x^2-7 x+5=0$

Divide by 2x on both the sides,

x + \(\frac{5}{2x}\) = \(\frac{7}{2}\)

Now, $x^3+\frac{125}{8 x^3}$ = (\(\frac{7}{2}\))³ - 3 × \(\frac{7}{2}\) × \(\frac{5}{2}\) 

$x^3+\frac{125}{8 x^3}$ = (\(\frac{343}{8}\)) - (\(\frac{105}{4}\))

$x^3+\frac{125}{8 x^3}$ =  (\(\frac{343 - 210}{8}\)) 

$x^3+\frac{125}{8 x^3}$ = $16 \frac{5}{8}$