Answer :

We have,

Applying R1 R1 – R2, we get






Taking (a + b+ c) common from first row, we get



Applying R2 R2 – R3, we get




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



Applying C1 C1 + C2 + C3, we get




Now, expanding along C1, we get


= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(c – b)(b – a) – (a – c)2}]


= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(cb – ac – b2 + ab – (a + c2 – 2ac)}]


= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){(cb – ac – b2 + ab – a - c2 + 2ac)}]


= (a + b + c)2[ab + bc + ca – (a2 + b2 + c2){ac + bc + ab – (a2 + b2 + c2)}]


=(a + b + c)2[ab + bc + ca – (a2 + b2 + c2)]2


=(a + b + c)(a + b + c)[ab + bc + ca – (a2 + b2 + c2)


Hence, given determinant is divisible by (a + b + c) and Quotient is (a + b + c)[ab + bc + ca – (a2 + b2 + c2)


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