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Using properties of determinants, prove the following: `|a^2a b a c a bb^2+1b cc a c b c^2+1|=1+a^2+b^2+c^2dot`

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`|{:(a^(2)+1,ab,ac),(ab,b^(2)+1,bc),(ca,cb,c^(2)+1):}|=abc|{:(a+1/a,b,c),(a,b+1/b,c),(a,b,c+1/c):}|`
`|{:(a^(2)+1,ab,ac),(ab,b^(2)+1,bc),(ca,cb,c^(2)+1):}|=|{:(a^(2)+b^(2)+c^(2)+1,b^(2),c^(2)),(a^(2)+b^(2)+c^(2)+1,b^(2)+1,c^(2)),(a^(2)+b^(2)+c^(2)+1,b^(2),c^(2)+1):}|`
`(C_(1)toC_(1)+C_(2)+C_(3))`
`=(a^(2)+b^(2)+c^(2)+1)|{:(1,b^(2),c^(2)),(1,b^(2)+1,c^(2)),(1,b^(2),c^(2)+1):}|`
`=(a^(2)+b^(2)+c^(2)+1)|{:(1,b^(2),c^(2)),(0,1,0),(0,0,1):}|`
`(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))`
`=(1+a^(2)+b^(2)+c^(2)).1|{:(1,0),(0,1):}|`
(Expanding along `C_(1)`)
`=(1+a^(2)+b^(2)+c^(2))=R.H.S.`
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