The Fourier sine series of the function f(x) = x³ on the interval [0,T] is 32(-1)+1 (nπ)²-6 -sin(nx). The solution of the equation m-1 is equal to? (a) u(x,t) = e-toz³ θα J²u = მე2 It u(x, 0) = 1³ u(0,t) = 0, u(π,t) = 0 (b) u(x,t) = 2(−1)÷1 (n)²–6 on²t sin(nx) (c) u(x,t)=2(−1)n−1 (nx)²–6¸ e sin(nx) ○ (d) u(x,t) = 2(-1)+1 (na)²=6e-on²t cos(nx) ○ (e) It's impossible to solve this equation

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