Prove the following identities –
= abc(a2 + b2 + c2)3
Let 
Taking a, b and c common from C1, C2 and C3, we get 
Recall that the value of a determinant remains same if we apply the operation Ri→ Ri + kRj or Ci→ Ci + kCj.
Applying R1→ R1 – R2, we get
Taking the term (a2 + b2 + c2) common from R1, we get
Applying R2→ R2 – R3, we get
Taking the term (a2 + b2 + c2) common from R2, we get
Applying C2→ C2 + C1, we get
Expanding the determinant along R1, we have
Δ = (abc)(a2 + b2 + c2)2(–1)[(a2 + b2 – c2) – (2b2 + 2a2)]
⇒ Δ = (abc)(a2 + b2 + c2)2[–(a2 + b2 – c2) + (2b2 + 2a2)]
⇒ Δ = (abc)(a2 + b2 + c2)2[–a2 – b2 + c2 + 2b2 + 2a2]
⇒ Δ = (abc)(a2 + b2 + c2)2[a2 + b2 + c2]
∴ Δ = (abc)(a2 + b2 + c2)3
Thus, 
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