By the Fundamental Theorem of Algebra, the polynomial \[ x^3+2x^2+7x+1 \] has three complex roots \(\alpha_1,\alpha_2,\alpha_3\), counted with multiplicity. Compute \[ \alpha_1^3+\alpha_2^3+\alpha_3^3. \]
Solution
1. The roots satisfy\[ \alpha_1+\alpha_2+\alpha_3=-2 \]and\[ \alpha_1\alpha_2 +\alpha_1\alpha_3 +\alpha_2\alpha_3 = 7. \]
These are Vieta’s formulas for \[ x^3+2x^2+7x+1 = (x-\alpha_1)(x-\alpha_2)(x-\alpha_3). \]
2. One has\[ \alpha_1^2+\alpha_2^2+\alpha_3^2=-10. \]
Squaring the first identity in step 1 gives \[ (\alpha_1+\alpha_2+\alpha_3)^2 = \alpha_1^2+\alpha_2^2+\alpha_3^2 +2(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3). \] Using both identities from step 1, \[ 4 = \alpha_1^2+\alpha_2^2+\alpha_3^2+14, \] which proves the claim.
3. Each root \(\alpha_i\) satisfies\[ \alpha_i^3 = -2\alpha_i^2-7\alpha_i-1. \]
Since \(\alpha_i\) is a root of \[ x^3+2x^2+7x+1, \] one has \[ \alpha_i^3+2\alpha_i^2+7\alpha_i+1=0. \] Rearranging gives the displayed identity.
4. The required sum is\[ \boxed{ \alpha_1^3+\alpha_2^3+\alpha_3^3=31 }. \]
Summing the identity in step 3 over \(i=1,2,3\) gives \[ \begin{aligned} \alpha_1^3+\alpha_2^3+\alpha_3^3 &= -2(\alpha_1^2+\alpha_2^2+\alpha_3^2) -7(\alpha_1+\alpha_2+\alpha_3) -3\\ &= -2(-10)-7(-2)-3\\ &= 31, \end{aligned} \] using steps 1 and 2.
5. Q.E.D.
Step 4 gives the requested value.