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In my last question which was Proving $x=\sqrt{a}+\sqrt{b}$ is the key root to solve $x^4-2(a+b)x^2+(a-b)^2=0$ ,I could find the coefficients(were very easy) of fourth-degree equation, so I went to study the case which has key root $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ and I found the equation as follow

$$x^8-k_{1}x^6+k_{2}x^4-k_{3}x^2+k_{4}=0$$ The roots of this quation take the form $$x=\sqrt{a}+\sqrt{b}+\sqrt{c}$$ $$x=\sqrt{a}+\sqrt{b}-\sqrt{c}$$ $$x=\sqrt{a}-\sqrt{b}-\sqrt{c}$$ $$x=\sqrt{a}-\sqrt{b}+\sqrt{c}$$ $$x=-\sqrt{a}+\sqrt{b}+\sqrt{c}$$ $$x=-\sqrt{a}-\sqrt{b}+\sqrt{c}$$ $$x=-\sqrt{a}+\sqrt{b}-\sqrt{c}$$ $$x=-\sqrt{a}-\sqrt{b}-\sqrt{c}$$ When $a$, $b$, and $c$ are real numbers

I could find only the $k1=4(a+b+c)$.

By which method can I find the others coefficients

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If $x=\sqrt{a}+\sqrt{b}+\sqrt{c}$ then $(x-\sqrt{a})^2=(\sqrt{b}+\sqrt{c})^2$, that is, $\xi=\sqrt{\alpha}+\sqrt{\beta}$ with $\alpha=4ax^2$, $\beta=4bc$ and $\xi=x^2+a-b-c$.

The result in your previous question yields $\xi^4-2(\alpha+\beta)\xi^2+(\alpha-\beta)^2=0$, that is, $$(x^2+a-b-c)^4-8(ax^2+bc)(x^2+a-b-c)^2+16(ax^2-bc)^2=0.$$ This polynomial is also $$\prod_{\pm}(x\pm\sqrt{a}\pm\sqrt{b}\pm\sqrt{c}),$$ where the product is over the eight possible choices of the three $\pm$ signs.

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