If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and 
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a^2+b^2+c^2)+2(ab+bc+ca) = 49$
So $23+2(ab+bc+ca) = 49\Rightarrow (ab+bc+ca) = 13$
Now from $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31\Rightarrow \frac{(a+1)\cdot (b+1)+(b+1)\cdot (c+1)+(c+a)\cdot (a+1)}{(a+1)(b+1)(c+1)} = 31$
So $\displaystyle \frac{(ab+bc+ca)+2(a+b+c)+3}{1+(a+b+c)+(ab+bc+ca)+abc} = 31\Rightarrow \frac{13+2\cdot 7+3}{1+7+13+abc} = 31$
So $\displaystyle \frac{30}{21+abc} = 31\Rightarrow 21\times 31+31(abc) = 30\Rightarrow (abc) = \frac{30-21\times 31}{31}=-\frac{621}{31}$
Now How can I calculate $a^3+b^3+c^3$
Is there is any better method by which we can calculate $abc$
Help me
Thanks

You know their sum, their sum of products taken by two, and their product. Use Vieta's identities and the cubic formula. (Another approach would be by employing Newton's identities).
