I can't seem to get the correct answer for this question..
I'll show you step by step what I've done:
$\displaystyle \sqrt{1+\left(x^3-\frac1{4x^3}\right)^2} = \sqrt{1+\frac{(4x^6-1)^2}{16x^6}} = \frac1{4x^3} \cdot \sqrt{16x^6+(4x^6-1)^2}$ $\displaystyle = \frac1{4x^3} \cdot \sqrt{16x^6+16x^{12} - 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{16x^{12} + 8x^6 + 1} = \frac1{4x^3} \cdot \sqrt{(4x^6+1)^2} = $
$\displaystyle \boxed{\frac{4x^6+1}{4x^3}}$
The most common mistake in algebra calculations, $\displaystyle \sqrt{x^2} \neq x$, $\displaystyle \sqrt{x^2} = |x|$!
So you're right until $\displaystyle \frac{\sqrt{(4x^6+1)^2}}{\sqrt{(4x^3)^2}}$
It should be
$\displaystyle \frac{|4x^6+1|}{|4x^3|} = \left | \frac{4x^6+1}{4x^3} \right |$