Question

The value of sin⁶A + cos⁶A + 3sin²A cos²A is​

Answer 1

Answer:

sin^6A+cos^6A

(sin²A)³+(cos²A)³ ............[a^6=(a²)³]

(sin²A+cos²A)³-3sin²Acos²A.........[(a+b)³=a³+b³+3ab]

1³-3sin²Acos²A

1-3sin²Acos²A

Step-by-step explanation:

Hence proved, mark as brainlist and stay awesome

Answer 2

Answer:

1 – 3sin²Acos²A

Step-by-step explanation:

sin⁶A + cos⁶A

= (sin²A)³ + (cos²A)³

= (sin²A + cos²A)(sin⁴A – sin²Acos²A + cos⁴A)

= 1 (sin⁴A – sin^2Acos²A + cos⁴A + 2sin²Acos²A – 2sin²Acos²A)

= (sin²A + cos²A) – 3sin²Acos²A

= 1 – 3sin²Acos²A