Question

If $x^2+\frac{1}{x^2}=7$ and $x\ne 0$; find the value of :

$7x^3+8x-\frac{7}{x^3}-\frac{8}{x}$

• A. $\pm 53\sqrt{5}$
• B. $\pm 16\sqrt{5}$
• C. $\pm 22\sqrt{5}$
• D. $\pm 64\sqrt{5}$

Answer 1

The correct option is D $\pm 64\sqrt{5}$

$7x^3+8x-\frac{7}{x^3}-\frac{8}{x}$
$=7(x^3-\frac{1}{x^3})+8(x-\frac{1}{x})$
$=7(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)+8(x-\frac{1}{x})$
Now,
$(x-\frac{1}{x}{)}^2=x^2+\frac{1}{x^2}-2(x\left)\right(\frac{1}{x})$
$=7-2$
$=5$
$=>x-\frac{1}{x}=\pm \sqrt{5}$
Putting the values we get,
$=7(\pm \sqrt{5})(7+1)+8(\pm \sqrt{5})$
$=\pm 64\sqrt{5}$