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.
 A: Note that $$a^3+b^3=(a+b)(a^2\color{red}-ab+b^2).$$
Let $s=\sin \theta,c=\cos\theta$ as you did. Since
$$(s-c)^4=(s^2+c^2-2sc)^2=(1-2sc)^2=1-4sc+4s^2c^2,$$
$$(s+c)^2=s^2+c^2+2sc=1+2sc,$$
$$\begin{align}s^6+c^6&=(s^2+c^2)(s^4-s^2c^2+c^4)\\&=s^4-s^2c^2+c^4\\&=s^4+2s^2c^2-3s^2c^2+c^4\\&=(s^4+2s^2c^2+c^4)-3s^2c^2\\&=(s^2+c^2)^2-3s^2c^2\\&=1-3s^2c^2,\end{align}$$
we have
$$\begin{align}3(s-c)^4+6(s+c)^2+4(s^6+c^6)&=3(1-4sc+4s^2c^2)+6(1+2sc)+4(1-3s^2c^2)\\&=(3+6+4)+(-12+12)sc+(12-12)s^2c^2\\&=13.\end{align}$$
A: *

*$$
\begin{align}
(\sin x+\cos x)^2 &=\sin^2+\cos^2 +2\sin x \cos x \\
&= 1+ 2 \sin x \cos x\\
&=1+\sin 2x
\end{align}$$


*$$
\begin{align}
(\sin x+\cos x)^4 &= (\sin^2+\cos^2 +2\sin x \cos x)2 \\
&=(1+ 2 \sin x \cos x)^2\\
&=(1+\sin 2x)^2
\end{align}
$$


*$$
\begin{align}
(\sin^6 x+\cos^6 x) &=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cdot\cos^2x)\\
&=1\left((\sin^2x+\cos^2x)^2-2\sin^2x\cdot\cos^2x-\sin^2x\cdot\cos^2x\right)\\
&=(1-3\sin^2x\cdot\cos^2x)
\end{align}
$$
A: Note that


*

*$(\sin x - \cos x)^4= (1-\sin 2x)^2 = 1 -2\sin 2x + \sin^2 2x;$

*$(\sin x + \cos x)^2= 1 + \sin 2x;$

*$(\sin x + \cos x)^6 = (\sin^2 x + \cos^2 x)^ 3 -3\sin x \cdot \cos x (\sin^2 x + \cos^2 x) = 1-\dfrac{3}{4}\sin^2 2x.$
There fore 
$$ LHS = 3(1 -2\sin 2x + \sin^2 2x) + 6(1+\sin 2x) +4\left( 1-\dfrac{3}{4}\sin^2 2x\right)-13=0$$

A: Alternative Simplification of $$\displaystyle \sin^6(x)+\cos^6(x)$$
Using $$\sin^2(x)+\cos^2(x)=1\Rightarrow \underbrace{\sin^2 x}_{a}+\underbrace{\cos^2 x}_{b}+\underbrace{(-1)}_{c}=0$$
Now Using The Formula, If $${a+b+c=0\;,}$$ Then $$a^2+b^2+c^3=3abc.$$
So $$\left(\sin^2x\right)^3+\left(\cos^2x\right)^3+(-1)^3=3\cdot \left(\sin^2x\right)\cdot \left(\cos^2x\right)\cdot (-1)$$
So $$\displaystyle \sin^6 x+\cos^6 x=1-3\sin^2 x\cdot \cos^2 x$$
A: 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)

