Using properties of determinants prove that:\(\begin{vmatrix}1& b+c &

1.	Using properties of determinants prove that:\(\begin{vmatrix}1& b+c & b^2+c^2 \\[0.3em]1& c+a &c^2+a^2\\[0.3em]1 & a+b& a^2+b^2\end{vmatrix}\) = (a-b)(b-c)(c-a)\|(1,b + c,b^2 + c^2)(1,c + a,c^2 + a^2)(1, a + b, a^2 + b^2)\| = (a-b)(b-c)(c-a)
Answer» \(\begin{vmatrix} 1& b+c & b^2+c^2 \\[0.3em] 1& c+a &c^2+a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) = \(\begin{vmatrix} 0& b-c & b^2-c^2 \\[0.3em] 0& c-a &c^2-a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃] = \(\begin{vmatrix} 0& b-c & (b-a)(b+a) \\[0.3em] 0& c-a &(c-b)(c+b)\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) = (b - a)(c - b)\(\begin{vmatrix} 0& 1 & b+a \\[0.3em] 0& 1 &c+b\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) [R₁’ = R₁/(b - a) & R₂’ = R₂/(c - b)] = (b - a)(c - b)[0 + 0 + 1{(c + b) - (b + a)}][expansion by first column] = (a - b)(b - c)(c - a)

Using properties of determinants prove that:\(\begin{vmatrix}1& b+c & b^2+c^2 \\[0.3em]1& c+a &c^2+a^2\\[0.3em]1 & a+b& a^2+b^2\end{vmatrix}\) = (a-b)(b-c)(c-a)|(1,b + c,b^2 + c^2)(1,c + a,c^2 + a^2)(1, a + b, a^2 + b^2)| = (a-b)(b-c)(c-a)

Answer»

\(\begin{vmatrix} 1& b+c & b^2+c^2 \\[0.3em] 1& c+a &c^2+a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)

= \(\begin{vmatrix} 0& b-c & b^2-c^2 \\[0.3em] 0& c-a &c^2-a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

= \(\begin{vmatrix} 0& b-c & (b-a)(b+a) \\[0.3em] 0& c-a &(c-b)(c+b)\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)

= (b - a)(c - b)\(\begin{vmatrix} 0& 1 & b+a \\[0.3em] 0& 1 &c+b\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) [R₁’ = R₁/(b - a) & R₂’ = R₂/(c - b)]

= (b - a)(c - b)[0 + 0 + 1{(c + b) - (b + a)}][expansion by first column]

= (a - b)(b - c)(c - a)