1.

Prove that |{:(1,,beta gamma+alpha delta,,beta^(2) gamma^(2)+alpha^(2),delta^(2)),(1,,gamma alpha+beta delta,,gamma^(2)alpha^(2)+beta^(2)delta^(2)),(1,,alpha beta + gamma delta ,,alpha^(2)beta^(2)+gamma^(2) delta^(2)):}|

Answer»

Solution :`|{:(1,,beta gamma+alpha DELTA,,beta^(2) gamma^(2)+alpha^(2),delta^(2)),(1,,gamma alpha+beta delta,,gamma^(2)alpha^(2)+beta^(2)delta^(2)),(1,,alpha beta + gamma delta ,,alpha^(2)beta^(2)+gamma^(2) delta^(2)):}|`
APPLYING`R_(1) to R_(1) -R_(2) " and" R_(2) to R_(2) -R_(3)`
`Delta = |{:(0,,beta(gamma -delta)+alpha (delta -gamma),,beta^(2)(gamma^(2)-delta^(2))+alpha^(2)(delta^(2)-gamma^(2))),(0,,gamma(alpha-delta)++beta (delta-alpha),,gamma^(2)(alpha^(2)-delta^(2))+beta^(2)(delta^(2)-alpha^(2))),(1,,alpha beta +gamma delta ,,alpha^(2) beta^(2) +gamma^(2) delta^(2)):}|`
`=(gamma -delta )(alpha -delta)|{:(0,,beta -alpha,,beta^(2)(gamma +delta)-alpha^(2)(gamma +delta)),(0,,gamma - beta ,,gamma^(2) (alpha+delta)),(1,,alpha beta + gamma delta ,,alpha^(2) beta ^(2) + gamma^(2) delta^(2)):}|`
` |{:(0,,beta-alpha,,(beta^(2)-alpha^(2))(gamma+delta)),(0,,gamma-beta,,gamma^(2)(alpha+delta)-beta^(2)(alpha+delta)),(1,,alphabeta+gammadelta,,alpha^(2)beta^(2)+gamma^(2)delta^(2)):}|`
`= (gamma -delta )(alpha-delta)(beta-alpha)(gamma-beta)|{:(0,,1,,(beta+alpha)(gamma+delta)),(0,,1,,(gamma+beta)(alpha+delta)),(1,,alpha beta+gamma delta,,a^(2)beta^(2)+gamma^(2)delta^(2)):}|`
`|{:(1,,(beta+alpha)(gamma+delta)),(1,,(gamma+beta)(alpha+delta)):}|`
`=(gamma - delta(alpha - delta)(beta-alpha)(gamma-beta)(gammadelta+alpha beta - beta gamma - alpha delta)`
`=(gamma-delta) (alpha-delta)(beta-alpha)(gamma-beta)(beta-delta)(alpha-gamma)`
`=-(alpha-delta)(beta-delta)(gamma-delta)(alpha-delta)(beta -gamma) (gamma - alpha)`


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