Extend the identity for cube of binomial to cube of trinomial

(a+b)^3=a^3+b^3+3ab(a+b)

Replace b with b+c, then

(a+b+c)^3 = a^3+(b+c)^3+3a(b+c)(a+b+c)

Expand the term of (b+c)^3

(a+b+c)^3 = a^3+b^3+c^3+3bc(b+c)+3a(b+c)(a+b+c)

Apply the given conditions

a+b+c = 0 and a^3+b^3+c^3=0

Then, the above equation is simplified to

0 = 3bc(b+c)

Therefore,

b = -c, a = 0

a^{15}+b^{15}+c^{15} = 0