Using properties of determinants, prove the following `|(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 determinant 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)

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)

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) .

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) .

Consider Delta=|(alpha,beta,gamma),(alpha^2,beta^2,gamma^2),(beta+gamma,gamma+alpha,alpha+beta)| Show that Delta=(alpha-beta)(beta-gamma)(gamma-alpha)(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)

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)

Evaluate: |[alpha, beta,gamma],[alpha^2,beta^2,gamma^2],[beta+gamma,gamma+alpha,alpha+beta]|

Prove that |[[alpha,beta,gamma],[alpha^2,beta^2,gamma^2],[alpha^3,beta^3,gamma^3]]|= alpha beta gamma (alpha-beta)(beta-gamma)(gamma-alpha)

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)