f(x)=ax3+bx2+cx+41f′(x)=3ax2+2bx+cx⇒f′(1)=3a+2b+c=2…(1) f′′(n)=6ax+2 b⇒f′′(1)=6a+2 b=43a+b=2……….(2) (1)−(2)b+c=0...(3) $\begin{aligned}
& \mathrm{f}(1)=40 \
& \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40
\end{aligned}use(3)a+41=40by(2)\begin{aligned}
& -3+b=2 \Rightarrow b=5 & c=-5 \
& a^2+b^2+c^2=1+25+25=51
\end{aligned}$