Evaluate $\displaystyle \small \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$.

Evaluate $\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$.

My solution:

First, note that $6x^4-12x^3-x+2$ can be factorized as $6(2)^4-12(2)^3-(2)+2=96-96-2+2=0$, factorize it using the long polynomial division by the factor $x-2$, we get $6x^4-12x^3-x+2=(x-2)(6x^3-1)$.

So evaluating $\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$ is equivalent in evaluating $\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$.

If we could think of another polynomial function of $f$, and multiply it to the top and bottom of the argument rational function inside the sixth root, which for certain the function of $f$ also has a factor of $x-2$, then we could cross out the $x-2$ from top and bottom. But this function of $f$ must fulfill another criterion, it has to be closely related to $\sqrt[3]{x^3-\sqrt{x^2+60}}$ and $\sqrt{x^2-\sqrt[3]{x^2+60}}$:

$\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}\cdot \frac{f(x)}{f(x)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

It's tempted to try out the option $f(x)=x^6-x^2-60$ since $f(2)=2^6-2^2-60=64-4-60=0$, and $f(x)=(x-2)(x+2)(x^4+4x^2+15)$

$\displaystyle \large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle =\large \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{(x-2)(6x^3-1)}{x+2}\cdot \frac{f(x)}{f(x)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle =\small \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{\cancel{(x-2)}(6x^3-1)}{x+2}\cdot \frac{x^6-x^2-60}{\cancel{(x-2)}(x+2)(x^4+4x^2+15)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{x+2}\cdot \frac{x^6-x^2-60}{(x+2)(x^4+4x^2+15)}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \sqrt[6]{x^6-x^2-60}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle \small = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^6-x^2-60}}{\sqrt[3]{x^6-x^2-60}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)\,\, \text{as $\frac{1}{6}=\frac{1}{2}-\frac{1}{3}$}$

$\displaystyle \small = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{(x^2)^3-\sqrt[3]{(x^2+60)^3}}}{\sqrt[3]{(x^3)^2-\sqrt{(x^2+60)^2}}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle =\small \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{(x^2-\sqrt[3]{x^2+60})}\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3-\sqrt{x^2+60}}\sqrt[3]{x^3+\sqrt{x^2+60}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle \small = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\cancel{\sqrt{(x^2-\sqrt[3]{x^2+60})}}\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\cancel{\sqrt[3]{x^3-\sqrt{x^2+60}}}\sqrt[3]{x^3+\sqrt{x^2+60}}}\cdot \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\cancel{\sqrt[3]{x^3-\sqrt{x^2+60}}}}{\cancel{\sqrt{x^2-\sqrt[3]{x^2+60}}}}\right)$

$\displaystyle = \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3+\sqrt{x^2+60}}}\right)$

Now, we can safely evaluate the limit by plugging in $x=2$ into the expression and get:

$\displaystyle \lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^4-12x^3-x+2}{x+2}}\cdot \frac{\sqrt[3]{x^3-\sqrt{x^2+60}}}{\sqrt{x^2-\sqrt[3]{x^2+60}}}\right)$

$\displaystyle =\lim_{{x}\to{2}}\left(\sqrt[6]{\frac{6x^3-1}{(x+2)^2(x^4+4x^2+15)}}\cdot \frac{\sqrt{x^4+x^2\sqrt[3]{x^2+60}+\sqrt[3]{(x^2+60)^2}}}{\sqrt[3]{x^3+\sqrt{x^2+60}}}\right)$

$\displaystyle =\sqrt[6]{\frac{6(2)^3-1}{(2+2)^2(2^4+4(2)^2+15)}}\cdot \frac{\sqrt{2^4+2^2\sqrt[3]{2^2+60}+\sqrt[3]{(2^2+60)^2}}}{\sqrt[3]{2^3+\sqrt{2^2+60}}}$

$\displaystyle =\sqrt[6]{\frac{47}{(4)^2(47)}}\cdot \frac{\sqrt{16+2^2(4)+(4)^2}}{\sqrt[3]{8+8}}$

$\displaystyle =\sqrt[6]{\frac{1}{4^2}}\cdot \frac{\sqrt{48}}{\sqrt[3]{16}}$

$\displaystyle =\frac{1}{2^{\frac{4}{6}}}\cdot \frac{4\sqrt{3}}{2^{\frac{4}{3}}}$

$\displaystyle =\frac{4\sqrt{3}}{4}$

$\displaystyle =\sqrt{3}$