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Answer 1

$\bigstar\;\;\sf{We\;know\;that : \boxed{\sf{(a + b)^3 = a^3 + b^3 + 3ab(a + b)}}}$

$\sf{Now, Consider : (sin^2\theta + cos^2\theta)^3}$

$\sf{\implies (sin^2\theta + cos^2\theta)^3= sin^6\theta + cos^6\theta + 3.sin^2\theta.cos^2\theta(sin^2\theta + cos^2\theta)}$

$\bigstar\;\;\sf{We\;know\;that : \boxed{\sf{sin^2\theta + cos^2\theta = 1}}}$

$\sf{\implies (1)^3 = sin^6\theta + cos^6\theta + 3.sin^2\theta.cos^2\theta(1)}$

$\sf{\implies sin^6\theta + cos^6\theta = 1 - 3.sin^2\theta.cos^2\theta\;-----\;[1]}$

$\bigstar\;\;\sf{We\;know\;that : \boxed{\sf{(a + b)^2 = a^2 + b^2 + 2ab}}}$

$\sf{Now, Consider : (sin^2\theta + cos^2\theta)^2}$

$\sf{\implies (sin^2\theta + cos^2\theta)^2= sin^4\theta + cos^4\theta + 2.sin^2\theta.cos^2\theta}$

$\sf{\implies (1)^2= sin^4\theta + cos^4\theta + 2.sin^2\theta.cos^2\theta}$

$\sf{\implies sin^4\theta + cos^4\theta = 1 - 2.sin^2\theta.cos^2\theta\;------\;[2]}$

$\sf{Given\;Question : 2(sin^6\theta + cos^6\theta) - 3(sin^4\theta + cos^4\theta) + 1}$

$\textsf{Substituting the Equations [1] and [2] in the given question, We get :}$

$\sf{\implies 2(1 - 3.sin^2\theta.cos^2\theta) - 3(1 - 2.sin^2\theta.cos^2\theta) + 1}$

$\sf{\implies 2 - 6.sin^2\theta.cos^2\theta - 3 + 6.sin^2\theta.cos^2\theta + 1}$

$\sf{\implies 2 - 3 + 1}$

$\sf{\implies 0}$

Answer 2

2(sin^6x + cos^6x) - 3(sin^4x + cos^4x)+1

=2[(sin^2x)^3+ (cos^2x)^3] - 3(sin^4x + cos^4x)+1

= 2(sin^2x + cos^2x)(sin^4x - sin^2xcos^2x + cos^4x)-3(sin^4x + cos^4x)+1 [a^3+b^3=(a+b)(a2-ab+b2)]

= 2(sin^4x - sin^2xcos^2x + cos^4x) - 3(sin^4x + cos^4x)+1 [As sin^2x + cos^2x=1]

= 2(sin^4x + cos^4x) - 2sin^2xcos^2x - 3(sin^4x + cos^4x)+1

= -(sin^4x + cos^4x) - 2sin^2xcos^2x+1

= -(sin^4x + cos^4x + 2sin^2xcos^2x)+1

= -[(sin^2x)^2+ (cos^2x)^2+ 2(sin^2x)(cos^2x)]+1

= -(sin^2x + cos^2x)^2+1 [As a2+b2+2ab=(a+b)2]

= -(1)^2+1 [As sin^2x + cos^2x=1]

= -1+1 =0

Hence, 2(sin^6x + cos^6x) - 3(sin^4x + cos^4x)+1=0

Hope it helps you