Q. 225.0( 1 Vote )

# Prove the following identities – = 2(a - b)(b - c)(c - a)(a + b + c)(CBSE 2015)

Let

Taking 2 common from C2, we get

Recall that the value of a determinant remains same if we apply the operation Ri Ri + kRj or Ci Ci + kCj.

Applying R2 R2 – R1, we get

Applying R3 R3 – R1, we get

We have the identity a3 – b3 = (a – b)(a2 + ab + b2)

Taking (b – a) and (c – a) common from R2 and R3, we get

We know that the sign of a determinant changes if any two rows or columns are interchanged.

By interchanging C1 and C2, we get

Expanding the determinant along C1, we have

Δ = –2(b – a)(c – a)(1)[(b2 + ba + a2) – (c2 + ca + a2)]

Δ = 2(a – b)(c – a)[b2 + ba + a2 – c2 – ca – a2]

Δ = 2(a – b)(c – a)[b2 + ba – c2 – ca]

Δ = 2(a – b)(c – a)[b2 – c2 + (ba – ca)]

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

Δ = 2(a – b)(c – a)(b – c)(b + c + a)

Δ = 2(a – b)(b – c)(c – a)(a + b + c)

Thus,

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