Question

please answer this question

Answer 1

Solution :

Now, $\left|\begin{array}{ccc}a^{2} & a^{2}-(b-c)^{2} & bc\\b^{2} & b^{2}-(c-a)^{2} & ca\\c^{2} & c^{2}-(a-b)^{2} & ab\end{array}\right|$

$\mathrm{C'_{2} \to C_{2}-C_{2}}$ gives

$=\left|\begin{array}{ccc}a^{2} & -(b-c)^{2} & bc\\b^{2} & - (c-a)^{2} & ca\\c^{2} & -(a-b)^{2} & ab\end{array}\right|$

Taking (- 1) common from $\mathrm{C_{2}}$ and $\mathrm{C'_{3}\to 2C_{3}}$ gives

$=-\frac{1}{2}\left|\begin{array}{ccc}a^{2} & b^{2}-2bc+c^{2} & 2bc\\b^{2} & c^{2}-2ca+a^{2} & 2ca\\c^{2} & a^{2}-2ab+b^{2} & 2ab\end{array}\right|$

$\mathrm{C'_{2}\to C_{2}+C_{3}}$ gives

$=-\frac{1}{2}\left|\begin{array}{ccc}a^{2} & b^{2}+c^{2} & 2bc\\b^{2} & c^{2}+a^{2} & 2ca\\c^{2} & a^{2}+b^{2} & 2ab\end{array}\right|$

$\mathrm{C'_{1}\to C_{1}+C_{2}}$ and taking 2 common from $\mathrm{C_{3}}$, we get

$=-\left|\begin{array}{ccc}a^{2}+b^{2}+c^{2} & b^{2}+c^{2} & bc\\a^{2}+b^{2}+c^{2} & c^{2}+a^{2} & ca\\a^{2}+b^{2}+c^{2} & a^{2}+b^{2} & ab\end{array}\right|$

Taking $\mathrm{(a^{2}+b^{2}+c^{2})}$ common from $\mathrm{C_{1}}$ gives

$\mathrm{=-(a^{2}+b^{2}+c^{2})\left|\begin{array}{ccc}1 & b^{2}+c^{2} & bc\\1 & c^{2}+a^{2} & ca\\1 & a^{2}+b^{3} & ab\end{array}\right|}$

$\mathrm{R'_{2}\to R_{2}-R_{1}}$ and $\mathrm{R'_{3}\to R_{3}-R_{2}}$ give

$\mathrm{=-(a^{2}+b^{2}+c^{2})\left|\begin{array}{ccc}1 & b^{2}+c^{2} & bc\\0 & a^{2}-b^{2} & ca-bc\\0 & a^{2}-c^{2} & ab-bc\end{array}\right|}$

Expanding along $\mathrm{R_{1}}$, we get

$\mathrm{=-(a^{2}+b^{2}+c^{2})[(a^{2}-b^{2})(ab-bc)-(a^{2}-c^{2})(ca-bc)]}$

$\mathrm{=-(a^{2}+b^{2}+c^{2})[b(a+b)(a-b)(a-c)-c(a-c)(a+c)(a-b)]}$

$\mathrm{=-(a^{2}+b^{2}+c^{2})(a-c)(a-b)[ab+b^{2}-ca-c^{2}]}$

$\mathrm{=-(a^{2}+b^{2}+c^{2})(a-c)(a-b)[(ab-ca)+(b^{2}-c^{2})]}$

$\mathrm{=-(a^{2}+b^{2}+c^{2})(a-c)(a-b)[a(b-c)+(b+c)(b-c)]}$

$\mathrm{=-(a^{2}+b^{2}+c^{2})(a-c)(a-b)(b-c)(a+b+c)}$

$\mathrm{=(a-b)(b-c)(c-a)(a+b+c)(a^{2}+b^{2}+c^{2})}$

$\to \boxed{\mathrm{\left|\begin{array}{ccc}a^{2} & a^{2}-(b-c)^{2} & bc\\b^{2} & b^{2}-(c-a)^{2} & ca\\c^{2} & c^{2}-(a-b)^{2} & ab\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)(a^{2}+b^{2}+c^{2})}}$

∴ " Option (4) " is correct.