Question

Prove:
${sin}^{6} x + {cos}^{6} x = 1 - 3 {sin}^{2}x \: {cos}^{2} x$
​

Answer 1

Step-by-step explanation:

sin^6 X + Cos^6 X

= (sin²X)³+(cos²X)³

= (sin²x+cos²x)³- 3 sin²x Cos²x (sin²x+cos²x)[here a³+b³= (a+b)²-3ab(a+b)]

= 1³-3sin²x cos²x ×1 (here according to law

sin² x+cos²x=1]

= 1 - 3sin²x cos²x(proved)

Answer 2

Step-by-step explanation:

sin^6 X + Cos^6 X

= (sin²X)³+(cos²X)³

= (sin²x+cos²x)³- 3 sin²x Cos²x (sin²x+cos²x)[here a³+b³= (a+b)²-3ab(a+b)]

= 1³-3sin²x cos²x ×1 (here according to law

sin² x+cos²x=1]

= 1 - 3sin²x cos²x(proved)