[tex]a^2+b^2+c^2=1[/tex]
Explanation
By definition.
[tex]\begin{gathered} (a+b+c)^2=a^2+b^2+c^2+2((ab)+(bc)+(ac)) \\ \end{gathered}[/tex]
so
Let
[tex]\begin{gathered} (a+b+c)=9 \\ ab+bc+ac=40 \\ \text{now, replace} \end{gathered}[/tex]
[tex]\begin{gathered} (a+b+c)^2=a^2+b^2+c^2+2((ab)+(bc)+(ac)) \\ 9^2=a^2+b^2+c^2+2(40) \\ 81=a^2+b^2+c^2+80 \\ \text{subtract 80 in both sides} \\ 81-80=a^2+b^2+c^2+80-80 \\ 1=a^2+b^2+c^2 \end{gathered}[/tex]
hence
[tex]a^2+b^2+c^2=1[/tex]
I hope this helps you
