Question

If $3x-\frac{4}{x}=4$ and $x\ne 0$; find : $27x^3-\frac{64}{x^3}$

• A. $208$
• B. $112$
• C. $167$
• D. $211$

Answer 1

The correct option is A $208$

$3x-\frac{4}{x}=4$

Squaring both the sides,we get,

$=>(3x-\frac{4}{x}{)}^2=(4{)}^2$

$=>(3x{)}^2+(\frac{4}{x}{)}^2-2\left(3x\right)\left(\frac{4}{x}\right)=16$

$=>9x^2+\frac{16}{x^2}-24=16$

$=>9x^2+\frac{16}{x^2}=16+24$

$=>9x^2+\left(\frac{16}{x^2}\right)=40$

Now,

$27x^3-\frac{64}{x^3}=(3x-\frac{4}{x})(9x^2+\frac{16}{x^2}+3x\times \frac{4}{x})$

$=\left(4\right)(40+12)$

$=\left(4\right)\left(52\right)$

$=208$