Maximum value of sin³a + sin³b + sin³c, when a+b+c = 180 degrees

Find the maximum value of sin³a + sin³b + sin³c, when a + b + c = 180 degrees. I'm self-taught and new to trig calculus, so please spell it out.

Thanks a ton!

z(a,b,c) = sin³a + sin³b + sin³c
G(a,b,c) = a+b+c
G = π


∇z = λ∇G
<3sin²(a)cos(a), 3sin²(b)cos(b), 3sin²(c)cos(c)> = λ<1,1,1>

So you need to solve these equations
3sin²(a)cos(a) = λ
3sin²(b)cos(b) = λ
3sin²(c)cos(c) = λ
a+b+c=π

You have four equations and four unknowns.

Thanks X, your answer helped me figure it out.

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well if a,b and c were each 60 degrees, then the value would be 3*(3sqrt3/8)=1.95
but if a=90,b=90, c=0, then the value would be 2, which i think is your best bet. sin 90= 1. so as theta approaches 180 or zero, sin theta approaches zero. so multiplying a number less than 1 to a power will make that number even smaller.