2 ( sin6A - cos6A ) - 3 ( sin4A + cos4A ) +4 ( sin2A + cos2A )
2(sin6x + cos6x) – 3(sin4x + cos4x) + 4(sin2x + cos2x)
We know that a3 + b3 = (a + b) (a2 – ab + b2)
2(sin6x + cos6x) – 3(sin4x + cos4x) + 4(sin2x + cos2x)
= 2(sin2x + cos2x)(cos4x – cos2x sin2x + sin4x) – 3(sin4x + cos4x) + 4
= 2(cos4x + sin4x – cos2x sin2x ) – 3[(sin2x + cos2x)2 – 2cos2x sin2x] + 4
= 2[(cos2x + sin2x)2 – 2cos2x sin2x – cos2x sin2x] – 3[ 1 – 2cos2x sin2x] + 4
= 2[1 – 3cos2x sin2x] – 3[1 – 2cos2x sin2x] + 4
= 2 – 6cos2x sin2x – 3 + 6cos2x sin2x + 4
= 3