If m= sin⁶x + cos⁶x, then m belongs to the interval

(a) [7/8, 5/4] (b) [1/2 , 5/8] (c) [1/4, 1]

Explain pls...

m=sin⁶x+cos⁶x=(sin²x+cos²x)(sin⁴x+cos⁴x-sin²xcos²x)

as a³+b³=(a+b)(a²-ab+b²)

we know that,

sin²x+cos²x=1

m=((sin²x+cos²x)²-3sin²xcos²x)

m=1-3sin²xcos²x

m=1-3/4(sin2x)²

we know 0<=sin²A<=1,

so -1<=-sin²2x<=0 multiply with -1,

-3/4<=-3/4(sin²2x)<=0 multiply with 3/4,

1/4<=1-3/4(sin²2x)<=1 add 1,

1/4<=m<=1

therefore "m" belongs to[1,1/4].