If a,b b,c, and c,a are the roots of `x^(2)-4x+3=0,x^(2)-8x+15=0` and `x^(2)-6x+5=0,`
`[{:(a^(2),+c^(2),a^(2)+b^(2)),(b^(2),+c^(2),a^(2)+c^(2)):}]+[{:(2ac,-2ab),(-2bc,-2ac):}]`

Explanation

`therefore" " x^(2)-4x+3=0`
`rArr" "(x-1)(x-3)=0 " " therefore x=1,3`
`x^(2)-8x+15=0`
`rArr " " (x-3)(x-5)=0 " "therefore x=3,5`
and `x^(2)-6x+5=0`
`rArr" " (x-5)(x-1)=0 " " therefore x=5,1`
it is clear that a = 1,3 and c=5
Now, `[(a^(2)+c^(2),a^(2)+b^(2)),(b^(2)+c^(2),a^(2)+c^(2))]+[(2ac,-2ab),(-2bc,-2ac)]`
`[(a^(2)+c^(2)+2ac,a^(2)+b^(2)-2ab),(b^(2)+c^(2)-2bc,a^(2)+c^(2)-2ac)]+[(2ac,-2ab),(-2bc,-2ac)]`
`[((1+5)^(2),(1-3)^(2)),((3-5)^(2),(1-5)^(2))]=[(36,4),(4,16)]`