Prove that `|[1,alpha,alpha^2+betagamma],[1,beta,beta^2+gammaalpha],[1,gamma,gamma^2+alphabeta]|`=`2(alpha-beta)(beta-y)(y-alpha)`

Using properties of determinants, prove that |[alpha, alpha^2, beta+gamma],[beta,beta^2,gamma+alpha],[gamma,gamma^2,alpha+beta]|=(beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Using properties of determinants, prove that: |[alpha,alpha^2,beta+gamma],[beta,beta^2,gamma+alpha],[gamma,gamma^2,alpha+beta]| = (beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Using properties of determinants. Prove that |(alpha,alpha^2,beta+gamma),(beta,beta^2,gamma+alpha),(gamma,gamma^2,alpha+beta)|=(beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Prove that: |[alpha,beta,gamma],[alpha^2,beta^2,gamma^2],[beta+gamma,gamma+alpha,alpha+beta]|=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma) .

Using properties of determinants in Exercise 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)(alpha-beta)

Prove that |[alpha,beta,gamma] ,[alpha^2,beta^2,gamma^2] , [beta+gamma, gamma+alpha, beta+alpha]| = (alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)

Without expanding the determinant, prove that: (i) |{:(alpha, alpha^(2), beta gamma),(beta, beta^(2), gamma alpha),(gamma, gamma^(2), alpha beta):}| =|{:(1,alpha^(2), alpha^(3)),(1, beta^(2), beta^(3)),(1, gamma^(2), gamma^(3)):}|

Prove the following : |{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=|{:(alpha,beta, gamma),(alpha^(2),beta^(2),gamma^(2)),(beta+gamma,gamma+alpha, alpha+beta):}|=(beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Show that |[1,alpha,alpha^2],[1,beta,beta^2],[1,gamma,gamma^2]|=(alpha-beta)(beta-gamma)(gamma-alpha)