Solving a system of algebraic equations with Maple

Question

I want to solve the following algebraic equation system using solve command. But, it gives only the trivial solution. I want to find a,b and c in terms of k. k is a constant here. Thanks in advance.

Input:

solve({-6*c+(3/2)*c^2-2*b-3*b*c+(3/2)*b^2-3*a*c+(k^2)*b-b+(3/2)*(a^2)+(k^2)*a-a=0,
-2*b-3*b*c+3*(b^2)-6*a*c+2*(k^2)*c-2*c-9*a*b+3*(k^2)*b-3*b+6*(a^2)+4*(k^2)*a-4*a=0,
(3/2)*(b^2)-3*a*c+(k^2)*c-c-9*a*b+3*(k^2)*b-3*b+9*(a^2)+6*(k^2)*a-a=0, 
-3*a*b+(k^2)*b-b+6*(a^2)+4*(k^2)*a-4*a=0, 
(3/2)*(a^2)+(k^2)*a-a=0},{a,b,c});

Output:

{a=0, b=0, c=0}

Answer 1

On Maple 2018.1. Add option to solve: allsolutions = true

solve({3/2*(a^2)+k^2*a-a = 0, -6*c+3/2*(c^2)-2*b-3*b*c+3/2*(b^2)-3*a*c+k^2*b- 
b+3/2*(a^2)+k^2*a-a = 0, 3*b*k^2+2*c*k^2-9*a*b-6*a*c+3*b^2-3*b*c-2*b-2*c- 
3*b+6*a^2+4*k^2*a-4*a = 0, 3/2*(b^2)-3*a*c+k^2*c-c-9*a*b+3*k^2*b- 
3*b+9*a^2+6*k^2*a-a = 0, 4*a*k^2+b*k^2+6*a^2-3*a*b-4*a-b = 0}, {a, b, c, k}, 
allsolutions = true)

#{a = 0, b = 0, c = 0, k = k}, {a = 0, b = 0, c = 4, k = 1}, {a = 0, b = 0, c = 4, k = -1}

Mathematica 11.3 give the same answer:

Answer 2

If you have a set of eqautions eqs in terms of the dependent names a,b,c,k then the call solve(eqs, {a,b,c}) is asking for solutions in which the remaining name k is "free".

You can instead call solve(eqs, {a,b,c,k}) and obtain these solutions.

restart;

eqs := {-6*c+(3/2)*c^2-2*b-3*b*c+(3/2)*b^2-3*a*c+(k^2)*b-b+(3/2)*(a^2)+(k^2)*a-a=0,
        -2*b-3*b*c+3*(b^2)-6*a*c+2*(k^2)*c-2*c-9*a*b+3*(k^2)*b-3*b+6*(a^2)+4*(k^2)*a-4*a=0,
        (3/2)*(b^2)-3*a*c+(k^2)*c-c-9*a*b+3*(k^2)*b-3*b+9*(a^2)+6*(k^2)*a-a=0, 
        -3*a*b+(k^2)*b-b+6*(a^2)+4*(k^2)*a-4*a=0, 
        (3/2)*(a^2)+(k^2)*a-a=0}:

solve(eqs, {a,b,c,k});

    {a = 0, b = 0, c = 0, k = k}, {a = 0, b = 0, c = 4, k = 1}, 
    {a = 0, b = 0, c = 4, k = -1}

That provides two additional solutions in which k was not free.