Using peoperties of determinants in questions 11 to 15, prove that :
`|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=(beta-gamma)(gamma-alpha)(alpha-beta+gamma)`

Explanation

L.H.S.=`|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=|{:(alpha,alpha^(2),alpha+beta+gamma),(beta,beta^(2),alpha+beta+gamma),(gamma,gamma^(2),alpha+beta+gamma):}|`
`(C_(3)toC_(3)+C_(1))`
`=(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta,beta^(2),1),(gamma,gamma^(2),1):}|`
`(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta-alpha,beta^(2)-alpha^(2),0),(gamma-alpha,gamma^(2)-alpha^(2),0):}|`
`(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))`
`(alpha+beta+gamma)(beta-alpha)(gamma-alpha)|{:(alpha,alpha^(2),1),(1,beta-alpha,0),(1,gamma-alpha,0):}|`
`=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)1.|{:(1,beta+alpha),(1,gamma+alpha)`
`=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)(gamma+alpha-beta-alpha)`
`=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)`
=R.H.S.