The Fourier sine series of the function f(x) = x³ on the interval [0, π] is The solution of the equation 3 x³ = 2(-1)+1 n=1 (Nπ) 2 - 6 n3 -sin(nx). is equal to? ди a²u Ət მ2 u(x, 0) = x³ - u(0,t) = 0, u(T,t)=0 O (a) u(x,t) = e -tox³ ○ (b) u(x,t) = 2Σx²-1 (−1) n+1 (nπ)² - -6 -on²t e n3 sin(nx) n=1 ○ (c) u(x,t) = 2 (−1) n+1 (nm) 2-6 sin(nx) Σx-1 n3 ○ (d) u(x,t) = 2x-1 (−1) n+1 (nm) 2-6 e-on²+ cos(nx) n=1 n³ (e) It's impossible to solve this equation

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