Fórmulas útiles:
$$ a^3+b^3=(a+b)\,(a^2-ab+b^2),\qquad \sin^2\alpha+\cos^2\alpha=1, $$
$$ \sin^4\alpha+\cos^4\alpha=(\sin^2\alpha+\cos^2\alpha)^2-2\sin^2\alpha\cos^2\alpha =1-2\sin^2\alpha\cos^2\alpha. $$

Desarrollo:
Primero factorizamos $\sin^6\alpha+\cos^6\alpha$ como suma de cubos con
$a=\sin^2\alpha$, $b=\cos^2\alpha$:
$$ \sin^6\alpha+\cos^6\alpha =(\sin^2\alpha+\cos^2\alpha)\big(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha\big) =\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha, $$
pues $\sin^2\alpha+\cos^2\alpha=1$.

Entonces, el lado izquierdo de la identidad (LLI) es
$$ \begin{aligned} \text{LLI}&=2(\sin^6\alpha+\cos^6\alpha)-3(\sin^4\alpha+\cos^4\alpha)+1\\ &=2(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha)-3(\sin^4\alpha+\cos^4\alpha)+1\\ &=(2-3)(\sin^4\alpha+\cos^4\alpha)-2\sin^2\alpha\cos^2\alpha+1\\ &=-(\sin^4\alpha+\cos^4\alpha)-2\sin^2\alpha\cos^2\alpha+1. \end{aligned} $$
Sustituimos $\sin^4\alpha+\cos^4\alpha=1-2\sin^2\alpha\cos^2\alpha$:
$$ \text{LLI}=-(1-2\sin^2\alpha\cos^2\alpha)-2\sin^2\alpha\cos^2\alpha+1 =-1+2\sin^2\alpha\cos^2\alpha-2\sin^2\alpha\cos^2\alpha+1=0. $$

Resultado final: Se verifica
$$ 2\big(\sin^6\alpha+\cos^6\alpha\big)-3\big(\sin^4\alpha+\cos^4\alpha\big)+1=0. $$