Question

Cos^6C + Sin^6C=1- 3Sin^2C Cos^2C​

Answer 1

Answer:

by using=(a+b)^3 where a=cos^2C, b=sin^2C

• (cos^2C+sin^2C)^3=sin^6C+cos^6C+3Cos^C Sin^2C (sin^2C+cos^2C)
• (1)^3=sin^6C+cos^6C +3cos^2C sin^2C (1)
• 1-3cos^C sin^2C=cos^2C+sin^2C.