Q. 255.0( 1 Vote )

# The value of determinant A. a3 + b3 + c3B. 3 bcC. a3 + b3 + c3 – 3abcD. none of these

We have,

Applying C2 C2 + C3, we get

Taking (a + b + c) common from second column, we get

Applying C1 C1 – C3, we get

Expanding along first row, we get

= (a + b + c)[(-b){c – b} – (1){-c2 – (-ab)} + a{-c – (-a)}]

= (a + b + c)(-bc + b2 + c2 – ab – ac + a2)

= a(-bc + b2 + c2 – ab – ac + a2) + b(-bc + b2 + c2 – ab – ac + a2) + c(-bc + b2 + c2 – ab – ac + a2)

= -abc + ab2 + ac2 – a2b – a2c + a3 – b2c + b3 + bc2 – ab2 – abc + a2b – bc2 + b2c + c3 – abc – ac2 + a2c

= a3 + b3 + c3 – 3abc

Hence, the correct option is (c)

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