The determinant left| {begin{array}{*{20}{c}} | vedclass

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Question

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\{ab}&{{b^2} + {x^2}}&{bc}\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} ight|$.

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${x^4}$

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(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\{ab}&{{b^2} + {x^2}}&{bc}\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} ight|$.Multiply $C_1, C_2, C_3$ by $a, b, c$ respectively and divide the determinant by $abc$:$\Delta = \frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}{a({a^2} + {x^2})}&{ab^2}&{ac^2}\{a^2b}&{b({b^2} + {x^2})}&{bc^2}\{a^2c}&{b^2c}&{c({c^2} + {x^2})}\end{array}} ight|$.Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively:$\Delta = \frac{abc}{abc}\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{b^2}&{c^2}\{a^2}&{{b^2} + {x^2}}&{c^2}\{a^2}&{b^2}&{{c^2} + {x^2}}\end{array}} ight| = \left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{b^2}&{c^2}\{a^2}&{{b^2} + {x^2}}&{c^2}\{a^2}&{b^2}&{{c^2} + {x^2}}\end{array}} ight|$.Applying $C_1 	o C_1 + C_2 + C_3$:$\Delta = ({a^2} + {b^2} + {c^2} + {x^2})\left| {\begin{array}{*{20}{c}}1&{b^2}&{c^2}\1&{{b^2} + {x^2}}&{c^2}\1&{b^2}&{{c^2} + {x^2}}\end{array}} ight|$.Subtracting $R_1$ from $R_2$ and $R_3$:$\Delta = ({a^2} + {b^2} + {c^2} + {x^2})\left| {\begin{array}{*{20}{c}}1&{b^2}&{c^2}\0&{x^2}&{0}\0&{0}&{x^2}\end{array}} ight| = ({a^2} + {b^2} + {c^2} + {x^2}) \cdot x^4$.Thus,the determinant is divisible by ${x^4}$.

The determinant $\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} \right|$ is a divisor of:

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