prove that 1 b+C b2+c2 1 c+a c2+a2 = (a-b)(b-c)(c-a) 1 a+b a2+b2 - Maths - Determinants

Question

prove that
1 b+C b2+c2
1 c+a c2+a2 = (a-b)(b-c)(c-a)
1 a+b a2+b2

Priyanka Kedia · Accepted Answer

$Δ = | \begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 1 & c + a & c^{2} + a^{2} \\ 1 & a + b & a^{2} + b^{2} \end{matrix} |$

Applying $R_{2} - R_{1} a n d R_{3} - R_{1}$ , we have,

$Δ = | \begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 0 & a - b & a^{2} - b^{2} \\ 0 & a - c & a^{2} - c^{2} \end{matrix} | \Rightarrow Δ = | \begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 0 & a - b & (a - b) (a + b) \\ 0 & a - c & (a - c) (a + c) \end{matrix} | \Rightarrow Δ = (a - b) (a - c) | \begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 0 & 1 & (a + b) \\ 0 & 1 & (a + c) \end{matrix} | \Rightarrow Δ = (a - b) (a - c) | \begin{matrix} 1 & a + b \\ 1 & a + c \end{matrix} | \Rightarrow Δ = (a - b) (a - c) (a + c - a - b) \Rightarrow Δ = (a - b) (a - c) (c - b) \Rightarrow Δ = (a - b) (b - c) (c - a) H e n c e p r o v e d .$