If $2 x^2-7 x+5=0$, then what is the value of $x^3+\frac{125}{8 x^3}$ ?
Answer & explanation
Correct answer: option 2
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 × \(\frac{7}{2}\) × \(\frac{5}{2}\)
$x^3+\frac{125}{8 x^3}$ = (\(\frac{343}{8}\)) - (\(\frac{105}{4}\))
$x^3+\frac{125}{8 x^3}$ = (\(\frac{343 - 210}{8}\))
$x^3+\frac{125}{8 x^3}$ = $16 \frac{5}{8}$