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in Determinants by (15.2k points)

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)

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1 Answer

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by (15.7k points)
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\(\begin{vmatrix} 1& b+c & b^2+c^2 \\[0.3em] 1& c+a &c^2+a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)

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

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