Question

If $3x-\frac{4}{x}=4$ and $x\ne 0$; find the value of

$27x^3-\frac{64}{x^3}$

• A. $208$
• B. $324$
• C. $114$
• D. $546$

Answer 1

The correct option is A $208$

$3x-\frac{4}{x}=4$
Squaring both sides,
$(3x-\frac{4}{x}{)}^2=4^2$
$=>(3x{)}^2+(\frac{4}{x}{)}^2-2\left(3x\right)\left(\frac{4}{x}\right)=16$
$=>(3x{)}^2+(\frac{4}{x}{)}^2=16+24$
$=>(3x{)}^2+(\frac{4}{x}{)}^2=40$
Now,
$27x^3-\frac{64}{x^3}=(3x-\frac{4}{x})\left(\right(3x{)}^2+(\frac{4}{x}{)}^2+(3x\left)\right(\frac{4}{x}\left)\right)$
$=4(40+12)$
$=208$