Prove sin^6 a + cos^6 a =1-3sin^2 a cos^ a
1) sin^6 a + cos^6 a = (sin^2 a)^3 + (cos^2 a)^3
2) a^3 + b^3 = (a+b) (a^2 - ab + b^2)
3) (sin^2 a)^3 + (cos^2 a)^3 = (sin^2 a + cos^2 a) (sin^4 a - sin^2 a* cos^2 a) + cos<span>^4 a)
4) </span><span>(sin^2 a + cos^2 a) = 1
5 ) </span>sin^6 a + cos^6 a = sin^4 a + cos^4 a - sin^2 a* cos^2 a<span>
6) a</span>^4 + b^4 = (a^2 +b^2 )- 2 a^2 b^2
7) sin^4 a + cos^4 a -sin^2 a* cos^2 a =(sin^2 a + cos^2 a)-2sin^2 acos<span>^2 a
8) </span>sin^6 a + cos^6 a = 1- 2sin^2 a cos^2 a - <span>sin^2 a* cos^2 a
9) </span><span>sin^6 a + cos^6 a = 1- 3 sin^2 a cos^2 a</span>
12+4 because 12-(-4), you keep, change, and flip. that means you keep 12 change the - to a + and you flip -4 to +4
Answer:
250%
Step-by-step explanation:
