Prove the following identities :\(\begin{vmatrix}a & b& c \\[0.3em]a-b

1.	Prove the following identities :\(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\) = a3 + b3 + c3 + 3abc
Answer» \(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\) L.H.S = \(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\) Apply C₁→C₁ + C₂ + C₃ = (a + b + c) \(\begin{vmatrix}1 & b& c \\[0.3em]0 & b-c & c-a \\[0.3em]2 & c+a & a+b\end{vmatrix}\) Applying, R₃→R₃ – 2R₁ = (a + b + c) \(\begin{vmatrix}1 & b& c \\[0.3em]0 & b-c & c-a \\[0.3em]0 & c+a-2b & a+b-2c\end{vmatrix}\) = (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)] = a³+ b³ + c³ – 3abc As, L.H.S = R.H.S Hence, proved.

Prove the following identities :\(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\) = a3 + b3 + c3 + 3abc

Answer»

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

L.H.S = \(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\)

Apply C₁→C₁ + C₂ + C₃

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

Applying,

R₃→R₃ – 2R₁

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

= (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)]

= a³+ b³ + c³ – 3abc

As,

L.H.S = R.H.S

Hence, proved.