Answer :

C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

C2 → C2 - C3 (i.e. Replacing 2^{nd} column by subtraction of 2^{nd} and 3^{rd} column)

We have a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) and b^{3} - c^{3} = (b - c)(b^{2} + bc + c^{2})

Therefore taking (a - b) and (b - c) outside the determinant from 1^{st} and 2^{nd} column respectively

C1 → C1 - C2 (i.e. Replacing 1^{st} column by subtraction of 1^{st} and 2^{nd} column)

As a^{2} - c^{2}) = (a + c)(a - c) therefore taking (a - c) outside the determinant from 1^{st} column we get

.

Expanding the determinant along 1^{st} row

∴ LHS = (a - b)(b - c)(a - c)( - (a + b + c))

Adjusting the minus sign with (a - c)

∴LHS = (a - b)(b - c)(c - a)(a + b + c) = RHS

∴

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Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Using properties Mathematics - Board Papers

Prove the followiMathematics - Board Papers