Answer :
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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