Write the value of 2
Question:

Write the value of 2 (sin6 x + cos6 x) −3 (sin4 x + cos4 x) + 1.

Solution:

$2\left(\sin ^{6} x+\cos ^{6} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$

$=2\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$

$=2 \cdot 1\left(\sin ^{4} x+\cos ^{4} x-\sin ^{2} x \cdot \cos ^{2} x\right)-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$

$=2\left(\sin ^{4} x+\cos ^{4} x\right)-2 \sin ^{2} x \cdot \cos ^{2} x-3\left(\sin ^{4} x+\cos ^{4} x\right)+1$

$=-\left(\sin ^{4} x+\cos ^{4} x\right)-2 \sin ^{2} x \cdot c \operatorname{os}^{2} x+1$

$=-\left\{\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cdot \cos ^{2} x\right\}+1$

$=-\left(\sin ^{2} x+\cos ^{2} x\right)^{2}+1$

$=-1+1$

= 0