# Prove the followi

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 + a + 1) 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 = (a2 + a + 1)(1)[(1 – a2)(1 – a) – (a2 – a)(a – a2)]

Δ = (a2 + a + 1)(1 – a – a2 + a3 – a3 + a4 + a2 – a3)

Δ = (a2 + a + 1)(1 – a – a3 + a4)

Δ = (a2 + a + 1)(a4 – a3 – a + 1)

Δ = (a2 + a + 1)[a3(a – 1) – (a – 1)]

Δ = (a2 + a + 1)(a – 1)(a3 – 1)

Δ = (a – 1)(a2 + a + 1)(a3 – 1)

Δ = (a3 – 1)(a3 – 1)

Δ = (a3 – 1)2

Thus, Rate this question :

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Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Prove the followiMathematics - Board Papers