For monic x^3 + ax^2 + bx + c with roots r_i, give p1 = Σ r_i, p2 = Σ r_i^2, p3 = Σ r_i^3.
Solution
Correct Answer: Option A
Let the monic cubic be \(f(x)=x^{3}+ax^{2}+bx+c\) with roots \(r_{1},r_{2},r_{3}\). By Viète's relations
\[
e_{1}=r_{1}+r_{2}+r_{3}=-a,\qquad
e_{2}=r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}=b,\qquad
e_{3}=r_{1}r_{2}r_{3}=-c.
\]
Denote \(p_{k}=r_{1}^{k}+r_{2}^{k}+r_{3}^{k}\). Then
1) \(p_{1}=e_{1}=-a.\)
2) For \(p_{2}\) use the square of the sum:
\[
p_{2}=(r_{1}+r_{2}+r_{3})^{2}-2\sum_{i\]
Substituting \(e_{1}=-a,\ e_{2}=b\) gives
\[
p_{2}=(-a)^{2}-2b=a^{2}-2b.
\]
3) For \(p_{3}\) use Newton's identity for degree 3:
\[
p_{3}=e_{1}p_{2}-e_{2}p_{1}+3e_{3}.
\]
Substitute the known values:
\[
p_{3}=(-a)(a^{2}-2b)-b(-a)+3(-c)=-a^{3}+2ab+ab-3c=-a^{3}+3ab-3c.
\]
Thus
\[
p_{1}=-a,\qquad p_{2}=a^{2}-2b,\qquad p_{3}=-a^{3}+3ab-3c,
\]
which matches Option 1.
\[
e_{1}=r_{1}+r_{2}+r_{3}=-a,\qquad
e_{2}=r_{1}r_{2}+r_{2}r_{3}+r_{3}r_{1}=b,\qquad
e_{3}=r_{1}r_{2}r_{3}=-c.
\]
Denote \(p_{k}=r_{1}^{k}+r_{2}^{k}+r_{3}^{k}\). Then
1) \(p_{1}=e_{1}=-a.\)
2) For \(p_{2}\) use the square of the sum:
\[
p_{2}=(r_{1}+r_{2}+r_{3})^{2}-2\sum_{i
Substituting \(e_{1}=-a,\ e_{2}=b\) gives
\[
p_{2}=(-a)^{2}-2b=a^{2}-2b.
\]
3) For \(p_{3}\) use Newton's identity for degree 3:
\[
p_{3}=e_{1}p_{2}-e_{2}p_{1}+3e_{3}.
\]
Substitute the known values:
\[
p_{3}=(-a)(a^{2}-2b)-b(-a)+3(-c)=-a^{3}+2ab+ab-3c=-a^{3}+3ab-3c.
\]
Thus
\[
p_{1}=-a,\qquad p_{2}=a^{2}-2b,\qquad p_{3}=-a^{3}+3ab-3c,
\]
which matches Option 1.
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