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Explicație pas cu pas:

[tex]\left|\begin{array}{ccc}a+x&x&x\\x&b+x&x\\x&x&c+x\end{array}\right| =[/tex]

[tex]= (a + x) \cdot \left|\begin{array}{ccc}b+x&x\\x&c+x\end{array}\right| - x \cdot \left|\begin{array}{ccc}x&x\\x&c+x\end{array}\right| + x \cdot \left|\begin{array}{ccc}x&b+x\\x&x\end{array}\right|[/tex]

[tex]= (a + x)\Big[(b + x)(c + x) - {x}^{2} \Big] - x\Big[x(c + x) - {x}^{2} \Big] + x\Big[{x}^{2} - x(b + x)\Big] \\ [/tex]

[tex]= (a + x)\Big(bc + bx + cx \Big) - c{x}^{2} - b{x}^{2} \\ [/tex]

[tex]= abc + x(ab + bc + ac)[/tex]

[tex]abc + x(ab + bc + ac) = 0 \\ \implies \bf x = - \dfrac{abc}{ab + bc + ac} [/tex]

[tex]\left|\begin{array}{ccc}a+x&x&x\\x&b+x&x\\x&x&c+x\end{array}\right|=(a+x)(b+x)(c+x)+x \cdot x \cdot x+x \cdot x \cdot x-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x \cdot (b+x) \cdot x-x \cdot x \cdot (c+x)-(a+x) \cdot x \cdot x=[/tex]

                                    [tex]~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(ab+ax+bx+x^2)(c+x)+x^3+x^3-x^2b-x^3-\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x^2c-x^3-x^2a-x^3=abc+abx+acx+x^2a+\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+bcx+x^2b+x^2c+x^3+x^3+x^3-x^2b-x^3-\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-x^2c-x^3-x^2a-x^3=abc+abx+acx+bcx[/tex]

[tex]\left|\begin{array}{ccc}a+x&x&x\\x&b+x&x\\x&x&c+x\end{array}\right|=0\Rightarrow abc+abx+acx+bcx=0 \Rightarrow \\\\ \Rightarrow abx+acx+bcx=-abc \Rightarrow x(ab+ac+bc)=-abc \Rightarrow \boxed{x=-\cfrac{abc}{ab+ac+bc}}[/tex]