15. Using properties of determinants, prove the following `|[a,b,c],[a-b,b-c,c-a],[b+c,c+a,a+b]|`=`a^3+b^3+c^3-3abc`

15. Using properties of determinants, prove the following `|[a,b,c],[a

1.	15. Using properties of determinants, prove the following `\|[a,b,c],[a-b,b-c,c-a],[b+c,c+a,a+b]\|`=`a^3+b^3+c^3-3abc`
Answer» Let the value of the given determinant be `Delta`. Then, `Delta = \|{:(a, b, c),(a-b, b-c, c-a),(b+c, c+a, a+b):}\|` `= \|{:(a+b+c, " "b, " "c),(" "0, b-c, c-a),(2(a+b+c), c+a, a+b):}\| " "[C_(1) to (C_(1) + C_(2) + C_(3))]` `= (a+b+c)\|{:(1, " "b, " "c),(0, b-c, c-a),(2, c+a, a+b):}\| " "["taking(a+b+c) common from"C_(1)]` `= (a+b+c)\|{:(1, " "b, " "c),(0, b-c, c-a),(0, c+a-2b, a+b-2c):}\| " "[R_(3) to R_(3)-2R_(1)]` `= (a+b+c)1\|{:(b-c, c-a),(c+a-2b, a+b-2c):}\| " "["expanded by"C_(1)]` `= (a+b+c)\|{:(b-c, c-a),(a-b, b-c):}\| [R_(2) to R_(2) + R_(1)]` `=(a+b+c) [(b-c)^(2) - (a-b)(c-a)]` ` =(a+b+c)(a^(2) +b^(2) +c^(2)-ab-bc-ca)` `=(a^(3) +b^(3) + c^(3) -3abc)` Hence, `Delta = (a^(3) +b^(3) +c^(3) -3abc)`