Prove the following identities –
Let 
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
Applying R1→ R1 + R3, we get
Taking the term (a2 + b2 + 2ab) common from R1, we get
Applying C2→ C2 – C1, we get
Applying C3→ C3 – C1, we get
Expanding the determinant along R1, we have
Δ = (a + b)2 [(a2 – b2)(a2 – 2ab) – (b2 – 2ab)(2ab – b2)]
⇒ Δ = (a + b)2 [a4 – 2a3b – b2a2 + 2ab3 – 2ab3 + b4 + 4a2b2 – 2ab3]
⇒ Δ = (a + b)2 [a4 – 2a3b + 3a2b2 – 2ab3 + b4]
⇒ Δ = (a + b)2 [a4 + b4 + 2a2b2 – 2a3b – 2ab3 + a2b2]
⇒ Δ = (a + b)2 [(a2 + b2)2 – 2ab(a2 + b2) + (ab)2]
⇒ Δ = (a + b)2 [(a2 + b2 – ab)2]
⇒ Δ = [(a + b)(a2 + b2 – ab)]2
∴ Δ = (a3 + b3)2
Thus, 
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