AMC 12A 2024 (Problem 15)The roots of x3+2x2−x+3x^3+2x^2-x+3x3+2x2−x+3 are ppp, qqq, and rrr. What is the value of (p2+4)(q2+4)(r2+4)(p^2+4)(q^2+4)(r^2+4)(p2+4)(q2+4)(r2+4)?(A) 64\text{(A)}\;64(A)64(B) 75\text{(B)}\;75(B)75(C) 100\text{(C)}\;100(C)100(D) 125\text{(D)}\;125(D)125(E) 144\text{(E)}\;144(E)144Related TopicsToolkit 24 — Square of a sum (n terms)Toolkit 30 — Vieta's FormulaHints (6)Hint 1By Toolkit 30 — Vieta's Formula. Vieta's Formula: p+q+r=−2p+q+r=-2p+q+r=−2pq+pr+qr=−1pq+pr+qr=-1pq+pr+qr=−1pqr=−3pqr=-3pqr=−3Hint 2(p2+4)(q2+4)(r2+4)=64+16(p2+q2+r2)+4(p2q2+p2r2+q2r2)+p2q2r2(p^2+4)(q^2+4)(r^2+4)=64+16(p^2+q^2+r^2)+4(p^2q^2+p^2r^2+q^2r^2)+p^2q^2r^2(p2+4)(q2+4)(r2+4)=64+16(p2+q2+r2)+4(p2q2+p2r2+q2r2)+p2q2r2Hint 3p2q2r2=(pqr)2=(−3)2=9p^2q^2r^2=(pqr)^2=(-3)^2=9p2q2r2=(pqr)2=(−3)2=9Hint 4p2+q2+r2=(p+q+r)2−2(pq+pr+qr)=(−2)2−2(−1)=6p^2+q^2+r^2=(p+q+r)^2-2(pq+pr+qr)=(-2)^2-2(-1)=6p2+q2+r2=(p+q+r)2−2(pq+pr+qr)=(−2)2−2(−1)=6Hint 5p2q2+p2r2+q2r2=(pq+pr+qr)2−2pqr(p+q+r)=(−1)2−2(−3)(−2)=1−12=−11p^2q^2+p^2r^2+q^2r^2=(pq+pr+qr)^2-2pqr(p+q+r)=(-1)^2-2(-3)(-2)=1-12=-11p2q2+p2r2+q2r2=(pq+pr+qr)2−2pqr(p+q+r)=(−1)2−2(−3)(−2)=1−12=−11Hint 6(p2+4)(q2+4)(r2+4)=64+16(6)+4(−11)+9=125(p^2+4)(q^2+4)(r^2+4)=64+16(6)+4(-11)+9=125(p2+4)(q2+4)(r2+4)=64+16(6)+4(−11)+9=125Final Answer(D) 125