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{4.1+3}=4.4 4.1^3 2^2.1^3 1^2.1^3 2.3 (k ➖ 3k+2).
Square the given equation and rearrange to k^3-3^3=0, which has roots k_j=3[cos(2jπ/3)+i*sin(2jπ/3)],j=0,1,2, where i=√(-1).k_0=3, k_1=(3/2)(-1+i√3) and k_2=(3/2)(-1-i√3)
@@wes9627 You are amazing
{4.1+3}=4.4 4.1^3 2^2.1^3 1^2.1^3 2.3 (k ➖ 3k+2).
Square the given equation and rearrange to k^3-3^3=0, which has roots k_j=3[cos(2jπ/3)+i*sin(2jπ/3)],j=0,1,2, where i=√(-1).
k_0=3, k_1=(3/2)(-1+i√3) and k_2=(3/2)(-1-i√3)
@@wes9627 You are amazing