Answer
$\sqrt{1+\Big(x^{3}-\dfrac{1}{4x^{3}}\Big)^{2}}=\dfrac{4x^{6}+1}{4x^{3}}$
Work Step by Step
$\sqrt{1+\Big(x^{3}-\dfrac{1}{4x^{3}}\Big)^{2}}$
Evaluate $\Big(x^{3}-\dfrac{1}{4x^{3}}\Big)^{2}$ inside the root:
$\sqrt{1+\Big(x^{3}-\dfrac{1}{4x^{3}}\Big)^{2}}=...$
$...=\sqrt{1+\Big[x^{6}-2(x^{3})\Big(\dfrac{1}{4x^{3}}\Big)+\Big(\dfrac{1}{4x^{3}}\Big)^{2}\Big]}=...$
$...=\sqrt{1+\Big(x^{6}-\dfrac{1}{2}+\dfrac{1}{16x^{6}}\Big)}=...$
Simplify the expression inside the root:
$...=\sqrt{1+x^{6}-\dfrac{1}{2}+\dfrac{1}{16x^{6}}}=\sqrt{x^{6}+\dfrac{1}{16x^{6}}+\dfrac{1}{2}}=...$
$...=\sqrt{\dfrac{x^{6}(16x^{6})+1+8x^{6}}{16x^{6}}}=\sqrt{\dfrac{16x^{12}+8x^{6}+1}{16x^{6}}}=...$
Factor the numerator of the expression inside the root:
$...=\sqrt{\dfrac{(4x^{6}+1)^{2}}{16x^{6}}}=...$
Evaluate the root:
$...=\dfrac{4x^{6}+1}{4x^{3}}$
