Let $z$ be a complex number such that $ z = a+bi$ where $a,b \in \mathbb{R}$ and $i$ is the imaginary unit. Find the modulus of $z$ knowing that:$$a^3 = 3(1+ab^2)$$ $$b^3 = 3(a^2b-1)$$ My attempt: I decided to add the two equations to get $$a^3 +b^3 = 3 +3ab^2+3a^2b - 3$$ $$\Rightarrow (a+b)(a^2-ab+b^2) = 3ab^2 + 3a^2b $$ $$\Rightarrow (a+b)(a^2-ab+b^2) = 3ab(b + a)$$ Since $a+b \neq 0$, we can divide by $a+b$ to get $a^2-4ab+b^2 = 0$ .....$(3)$
Now I subtract the second equation from the first equation: $$a^3-b^3 = 3 + 3ab^2-3a^2b +3$$ $$\Rightarrow (a-b)(a^2+ab+b^2) = 6-3ab(a-b)$$ Assuming $a-b \neq 0$, We continue and get $$a^2+4ab+b^2 = 6....(4)$$ Now we add the third and fourth equations to get $$2a^2 + 2b^2 = 6$$ $$\Rightarrow a^2+b^2 = 3$$ $$ \Rightarrow \sqrt{a^2+b^2} = \sqrt{3}$$
However, when I checked the solutions, it said that the answer was $1.619$ (rounded off). Please can someone help me find my mistake?
I recommend one more way to solve it. $$z^3=(a^3-3ab^2)+(3a^2b-b^3)i=3+3i \Rightarrow |z|^3=3\sqrt{2} \Rightarrow |z|=\sqrt[6]{18}.$$