For the given expression prove that LHS is equal to the RHS. The expression is:
\[si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta \;\;=1\]

Hint: Using the identity \[si{{n}^{2}}\theta +co{{s}^{2}}\theta \;=1\], and the formula \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\] we can simplify the LHS and make it equal to the RHS.

Complete step-by-step answer:
In the given problem, we have to prove that \[si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta \;\;=1\]
So now we will start with the LHS.
\[\Rightarrow si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta \]
Now, it is know that \[si{{n}^{2}}\theta +co{{s}^{2}}\theta \;=1\], so we can write above expression as:
\[\begin{align}
  & \Rightarrow si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta \\
 & \Rightarrow si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta (si{{n}^{2}}\theta +co{{s}^{2}}\theta \;) \\
\end{align}\]
Now, let \[si{{n}^{2}}\theta =a\] and \[{{\cos }^{2}}\theta =b\], so we will have LHS as:
\[\begin{align}
  & \Rightarrow si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta (si{{n}^{2}}\theta +co{{s}^{2}}\theta \;) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+3ab(a+b) \\
 & \Rightarrow {{(a+b)}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)) \\
\end{align}\]
So substituting back the value of a and b, we have:
\[\begin{align}
  & \Rightarrow {{(a+b)}^{3}} \\
 & \Rightarrow {{(si{{n}^{2}}\theta +{{\cos }^{2}}\theta )}^{3}} \\
 & \Rightarrow {{(1)}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because si{{n}^{2}}\theta +co{{s}^{2}}\theta \;=1) \\
 & \Rightarrow 1 \\
 & = RHS \\
\end{align}\]

Hence, we have proved that LHS is equal to RHS in \[si{{n}^{6}}\theta +co{{s}^{6}}\theta \;+3\;si{{n}^{2}}\theta \,co{{s}^{2}}\theta \;\;=1\].

Hence proved.

Note: When we use the substitution of the trigonometric ratios, in order to simplify the given trigonometric expression, then it is important to substitute back and get the result in the original trigonometric ratios form. These types of problems can be solved using basic trigonometric and algebraic identities.