10. Let $\mathrm{f}$ be a real-valued function with the rule $\mathrm{f}(x)=x^{3}+3 x^{2}+6 x+14$ defined for all real value of $x$. It is given that $a$ and $b$ are two real numbers such that $\mathrm{f}(a)=1$ and $\mathrm{f}(b)=19$. Find the value of $(a+b)^{2}$.
10. Answer: 4
We note that $\mathrm{f}(x)=(x+1)^{3}+3(x+1)+10$. Let $\mathrm{g}(y)=y^{3}+3 y$, which is a strictly increasing odd function. Hence $\mathrm{f}(a)=1$ implies $\mathrm{g}(a+1)=-9$ and $\mathrm{f}(b)=19$ implies $\mathrm{g}(b+1)=9$. Thus, $a+1=-(b+1)$, implying $a+b=-2$, so that $(a+b)^{2}=4$.