If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$

Question

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

Answer 1

Note that $a,b,c$ are the roots of the equation $$x^3-(a+b+c)x^2+(bc+ac+ab)x-abc=0$$ which we know to be $$x^3-7x^2+13x-\frac {621}{31}=0 \dots (A)$$

Add the three equations for $a, b, c$ to obtain

$$(a^3+b^3+c^3)-7(a^2+b^2+c^2)+13(a+b+c)-\frac {3\cdot 621}{31}=0$$

Note that if we define $P_n=a^n+b^n+c^n$ we can multiply equation $A$ by $x^n$ before substituting $a,b,c$ and we get $$P_{n+3}-7P_{n+2}+13P_{n+1}-\frac {621}{31}P_n=0$$ which is a recurrence relation for the sums of higher powers. It works with negative powers too, provided the roots are all non-zero.

Answer 2

Let $p(x) = (x-a)(x-b)(x-c)$. Then the following identity holds:

$$ a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3p(a+b+c). $$

Since we know that $p(x) = x^{3}- 7 x^{2} + 13 x + \frac{621}{31}$, we now have the answer.

Answer 3

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).

Answer 4

HINT:

Where you have left of,

we can derive $$a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+b+ca)]$$

Now, $(a+b+c)^2-(a^2+b^2+c^2)=2(ab+bc+ca)$