Exercise 1117Algebra › PolynomialsL1Added 3 Nov 2026

Vieta's formulas for a cubic

#Vieta#cubic#symmetric functions

Statement

Let $a,b,c$ be the roots of $X^3-3X^2+2X-1=0$ . Compute $a+b+c$ , $ab+bc+ca$ , $abc$ , and $a^2+b^2+c^2$ .

Hint

For $X^3+pX^2+qX+r=0$ : sum $=-p$ , sum of pairwise products $=q$ , product $=-r$ .

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Vieta. For $X^3-3X^2+2X-1$ : $p=-3$ , $q=2$ , $r=-1$ .
$a+b+c=3,\qquad ab+bc+ca=2,\qquad abc=1.$

Sum of squares. $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-4=5$ .

Conclusion. $a+b+c=3$ , $ab+bc+ca=2$ , $abc=1$ , $a^2+b^2+c^2=5$ .