# Prove the following identities:2(sin6 x + cos6 x) – 3(sin4 x + cos4 x)+ 1 =0

To prove: 2(sin6 x + cos6 x) – 3(sin4 x + cos4 x)+ 1 = 0

Proof:

Take LHS:

2(sin6 x + cos6 x) – 3(sin4 x + cos4 x)+ 1

Identities used:

(a + b)2 = a2 + b2 + 2ab

a3 + b3 = (a + b) (a2 + b2 – ab)

Therefore,

= 2{(sin2 x)3 + (cos2 x)3} – 3{(sin2 x)2 + (cos2 x)2} + 1

= 2{(sin2 x + cos2 x)(sin4 x + cos4 x – sin2 x cos2 x} – 3{(sin2 x)2 + (cos2 x)2 + 2sin2 x cos2 x – 2sin2 x cos2 x }+ 1

= 2{(1)(sin4 x + cos4 x + 2 sin2 x cos2 x – 3 sin2 x cos2 x}–3{(sin2 x + cos2 x)2 – 2sin2 x cos2 x} + 1

{ sin2 x + cos2 x = 1}

= 2{(sin2 x + cos2 x)2 – 3 sin2 x cos2 x} – 3{(1)2 – 2sin2 x cos2 x } + 1

= 2{(1)2 – 3 sin2 x cos2 x} – 3(1 – 2sin2 x cos2 x) + 1

= 2(1 – 3 sin2 x cos2 x) – 3 + 6 sin2 x cos2 x + 1

= 2 – 6 sin2 x cos2 x – 2 + 6 sin2 x cos2 x

= 0

= RHS

Hence Proved

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