<s>[INST] Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$ 
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma

My Attempt::
For writing convenience, let $\color{red}{s} = \sin \theta$ and $\color{blue}{c} = \cos \theta$
$
\begin{align}
\text{To Prove }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) - 13 = 0\\
\equiv \text{TP }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) = 13
\end{align}
$
$$
\require{cancel}
\begin{align}
\text{LHS } 
&= 3\left[(\color{red}{s}-\color{blue}{c})^2\right]^2 
+ 6(\color{red}{s} +\color{blue}{c})^2
+ 4(\color{red}{s^6} + \color{blue}{c^6})
\\
&= 3(\color{limegreen}{s^2 + c^2} - 2\color{red}{s}\color{blue}{c})^2
+ 6(\color{limegreen}{s^2 + c^2} + 2\color{red}{s}\color{blue}{c})
+ 4\left[(\color{red}{s^2})^3 + (\color{blue}{c^2})^3\right]
\\
&= 3(1 - 2\color{red}{s}\color{blue}{c})^2 + 6(1+2\color{red}{s}\color{blue}{c}) 
+ 4(\color{limegreen}{s^2 + c^2})(\color{red}{s^4} 
- \color{red}{s^2}\color{blue}{c^2} 
+ \color{blue}{c^4})
\\
&= 3(1 - 4\color{red}{s}\color{blue}{c} + 4\color{red}{s^2}\color{blue}{c^2}) + 6 (1+2\color{red}{s}\color{blue}{c}) + 4(\color{red}{s^4} - \color{red}{s^2} \color{blue}{c^2}+\color{blue}{c^4})
\\
&= 3 
\cancel{- 12\color{red}{s}\color{blue}{c}} 
+12\color{red}{s^2}\color{blue}{c^2} 
+ 6 
\cancel{+ 12\color{red}{s}\color{blue}{c}} + 4\color{red}{s^4} 
-4\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4}\\
&= 4\color{red}{s^4} + 8\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4} + 9\\
&= 4(\color{red}{s^4} + 2\color{red}{s^2}\color{blue}{c^2} + \color{blue}{c^4}) + 9\\
&= 4(\color{limegreen}{s^2 + c^2})^2 + 9\\
&= 4 + 9 = 13 = \text{RHS} \tag{Q.E.D.}
\end{align}
$$
Thanks to @mathlove, I found and corrected the mistake in my attempt.
$\Huge\color{lightgrey}{☺}$ Although, a quicker alternate way will always be nice.
 [/INST] Here is another way to prove such identities:
Get the groebner basis of all the relevant polynomial equalities for what you are dealing with. In this case it's very simple since we only have $\sin\theta=s$ and $\cos\theta=c$ and the relevant polynomial equality is $s^2+c^2-1=0$. This is your groebner basis - no extra work required. However if you additionally had terms like $\sin2\theta=s_2$ and $\cos2\theta=c_2$ you would need $s_2^2+c_2^2-1=0$ and equalites that relate $s,c,s_2$ and $c_2$.
Now once you have this groebner basis reduce your polynomial (the equality you want to prove) by the groebner basis. This yields 0 if and only if the polynomial is in the ideal of the polynomial system. If it is a member of the ideal your equality is proved.
The nice thing about this approach is that you can do it in four lines of sympy code:
from sympy import *

s, c = symbols('s, c')

G = groebner([s**2+c**2-1])

print G.contains(3*(s-c)**4+6*(s+c)**2+4*(s**6+c**6)-13)

</s>