Prove that `|a a+b a+b+c2a3a+2B4a+3b+2c3a6a+3b 10 a+6b+3c|=a^3`

Explanation

Solving L.H.S.

`|[a, a+b, a+b+c],[2a,3a+2b,4a+3b+2c],[3a,6a+3b, 10 a+6b+3c]|`

Applying `R_ 2 to R_2-2R_1`

` " " " " =|[a, a+b, a+b+c],[2a-2(a),3a+2b-2(a+b),4a+3b+2c-2(a+b+c)],[3a,6a+3b, 10 a+6b+3c]|`

` " " " " =|[a, a+b, a+b+c],[0,3a+2b-2a-2b,4a+3b+2c-2a-2b-2c],[3a,6a+3b, 10 a+6b+3c]|`

` " " " " =|[a, a+b, a+b+c],[0,a,2a+b],[3a,6a+3b, 10 a+6b+3c]|`

Applying `R_ 3 to R_3-3R_1`

` " " " " =|[a, a+b, a+b+c],[0,a,2a+b],[3a-3(a),6a+3b-3(a+b), 10a+6b+3c-3(a+b+c)]|`

` " " " " =|[a, a+b, a+b+c],[0,a,2a+b],[0,6a+3b-3a-3b, 10a+6b+3c-3a-3b-3c]|`

` " " " " =|[a, a+b, a+b+c],[0,a,2a+b],[0,3a, 7a+3b]|`

Expanding along `C_1`

` " " " " =a|[a,2a+b],[3a,7a+3b]|-0|[a+b,a+b+c],[3a,7a+3b]|+0|[a+b,a+b+c],[3a,2a+b]|`

` " " " " =a|[a,2a+b],[3a,7a+3b]|-0+0`

` " " " " =a|[a,2a+b],[3a,7a+3b]|`

` " " " " = a[a(7a+3b)-3a(2a+b)]`

` " " " " = a[7a^2+3ab-6a^2-3ab]`

` " " " " = a(a^2)`

` " " " " = a^3`

` " " " " =` R.H.S.

L.H.S. = R.H.S.

Hence proved