Dirty Algebra

Let $a=\sin { \left( x \right) }$ , so $\sqrt { 1-{ a }^{ 2 } } =\cos { \left( x \right) }$

Now we have $\sin { \left( x \right) } \cos { \left( x \right) } =\frac { 1 }{ \sqrt { 6 } }$

Doubling both sides $2\sin { \left( x \right) } \cos { \left( x \right) } =\frac { 2 }{ \sqrt { 6 } }$

or $\sin { \left( 2x \right) } =\frac { 2 }{ \sqrt { 6 } }$

Now we need to find ${ a }^{ 6 }+{ \left( 1-{ a }^{ 2 } \right) }^{ 3 }$

Which is equal to ${ a }^{ 6 }+{ \left( \sqrt { 1-{ a }^{ 2 } } \right) }^{ 6 }$

or

$\sin ^{ 6 }{ \left( x \right) } +\cos ^{ 6 }{ \left( x \right) }$

Which is equal to (proof in comments section) $1-\frac { 3 }{ 4 } \sin ^{ 2 }{ \left( x \right) }$

Which gives $1-\frac { 3 }{ 4 } { \left( \frac { 2 }{ \sqrt { 6 } } \right) }^{ 2 }$

$=\quad \boxed { \frac { 1 }{ 2 } }$

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Although this looks quite long, by this method, it can be done mentally in seconds.

Please upvote if you liked the solution.

Squaring the given equation gives $(a\sqrt{1 - a^{2}})^{2} = (\frac{1}{\sqrt{6}})^{2}$ $a^{2} - a^{4} = \frac{1}{6}$

Going back to the question, let's factor \[\begin{array}{} a^{6} + (1- a^{2})^{3} & = (a^{2})^{3} + (1- a^{2})^{3} \\& = (a^{2} + 1 - a^{2})((a^{2})^{2} - a^{2}(1 - a^{2}) + (1 - a^{2})^{2}) \\& = (a^{2} +1 - a^{2})(a^{4} - a^{2} +a^{4} +1 - 2a^{2} + a^{4}) \\& = (1)(3a^{4} - 3a^{2} + 1) \\& = (1)(3(-\frac{1}{6}) + 1) \qquad \left[ \because a^{2} - a^{4} = \frac{1}{6} \right] \\& = \boxed {\frac{1}{2} \rightarrow 0.5} \end{array}\]

$\text{Here is a proof without trigonometry: }$

$a\sqrt{1-a^2}=\frac{1}{\sqrt{6}} \\ a^2(1-a^2)=\frac{1}{6} \\ a^2-a^4=\frac{1}{6} \\ \color{#3D99F6}{0=a^4-a^2+ \frac{1}{6}}$

$\text{Now let }a^6+(1-a^2)^3=S,$

$a^6-a^6+3a^4-3a^2+1=S \\ 3a^4-3a^2+1=S \\ \color{#D61F06}{a^4-a^2+\frac{1}{3}= \frac{S}{3}}$

$\text{and subtract the first equation from the second one:}$

$\color{#D61F06}{a^4-a^2+\frac{1}{3}}-\color{#3D99F6}{a^4+a^2-\frac{1}{6}}=\color{#D61F06}{\frac{S}{3}}-\color{#3D99F6}{0} \\ \frac{1}{6} = \frac{S}{3} \\ S = \large{\boxed{\frac{1}{2}}}$

Take $a = \sin \theta$

$\sin \theta \cos \theta = \frac{1}{\sqrt{6}}$

$\sin \theta \times \cos \theta = \frac{1}{\sqrt{6}}$

$\sin^6\theta + \cos^6\theta = (\sin^2\theta +\cos^2\theta)(\sin^4\theta - \sin^2\theta \cos^2\theta + \cos^4\theta)$

$= (1)((\sin^2\theta + \cos^2\theta)^2 - 2\sin^2\theta \cos^2\theta - \sin^2\theta \cos^2\theta) = 1 - 3\sin^2\theta \cos^2\theta$

$\Rightarrow a^6 + (1-a^2)^3 = (1)(1 - 3\times \frac{1}{6}) = \boxed{\color{#D61F06}{0.5}}$