Let `triangle =|[2a_1b_1, a_1b_2+a_2b_1, a_1b_3+a_3b_1], [a_1b_2+a_2b_1, 2a_2b_2, a_2b_3+a_3b_2], [a_1b_3+a_3b_1, a_3b_2+a_2b_3, 2a_3b_3]|` . Expressing `` as the product of two determinants, show that `triangle=0`

Explanation

`Delta = |{:(2a_(1)b_(1),,a_(1)b_(2)+a_(2)b_(1),,a_(1)b_(3)+a_(3)b_(1)),( a_(1)b_(2)+a_(2)b_(1),,2a_(2)b_(2),,a_(2)b_(3)+a_(3)b_(2)),(a_(2)b_(3)+a_(3)b_(1),,a_(3)b_(2)+a_(2)b_(3),,2a_(3)b_(3)):}|.`
`=|{:(a_(1),,b_(1),,0),( a_(2),,b_(2),,0),(a_(3),,b_(3),,0):}| xx |{:(b^(1),,a_(1),,0),( b_(2),,a_(2),,0),(b_(3),,a_(3),,0):}|=0`
Now `ax^(2) +2hxy +by^(2) +2gy +2fy+c`
`=(lx+ my+n) (l'x+m'y+n')`
`=ll'x^(2) +(lm +ml') xy+ mm'y^(2) +(ln'+l'n)x`
`+(mn'+m'n)y+nn'`
Comparing the coefficient we get
`a=ll' ,h =(1)/(2) (lm'+ml ),b =mm`
`g=(1)/(2) (ln' +nl) ,f =(1)/(2) (mn' +nn) ,c=nn`
`:. |{:(a,,h,,g),( h,,b,,f),(f,,f,,c):}|`
`=|{:(ll,,(1)/(2)(lm'+ml'),,(1)/(2)(ln'+nl')),((1)/(2)(lm'+l'n),,mm,,(1)/(2)(mn'+m'n)),((1)/(2)(ln'+l'n),,(1)/(2)(mn'+m'n),,n n):}|`
` =(1)/(8) |{:(2ll',,lm'+l'm,,ln'+l'n),( lm'+l'm,,2mm',,mn'+m'n),(ln'+l'n,,mn'+m'n,,2n n):}|`
`=0`