Let Δ=−bcb2+bcc2+bca2+ac−acc2+aca2+abb2+ab−ab and the equation px3+qx2+rx+s=0 has roots a, b, c, where a, b, c∈R+.The value of Δ isThe value of ∆ isIf Δ=27, and a2 + b2 + c2 = 3, then

Multiplying R1, R2, R3 by a, b, c, respectively, and then taking a, b, c common from C1, C2, and C3, we getΔ=−bcab+acac+abab+bc−acbc+abac+bcbc+ac−abNow, using C2→C2−C1 and C3→C3−C1, and then taking ( ab + bc + ca) common from C2 and C3, we getΔ=−bc11ab+bc−10ac+bc0−1×(ab+bc+ca)2Now, applying R2→R2+R1, we getΔ=−bc11ab01ac+bc0−1(ab+bc+ca)2Expanding along C2, we getΔ=(ab+bc+ca)2[ac+bc+ab]  =(ab+bc+ca)3  =(r/p)3=r3/p3Now given a, b, c are all positive, then A.M. ≥ G.M. ⇒ ab+bc+ac3≥(ab×bc×ac)1/3or   (ab+bc+ac)3≥27a2b2c2or   (ab+bc+ac)3≥27s2/p2If Δ=27, then ab + bc + ca = 3, and given that a2 + b2 + c2 = 3, from (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), we havea+b+c=±3⇒    a+b+c=3           ∵a, b, c∈R+⇒    3p+q=0Multiplying R1, R2, R3 by a, b, c, respectively, and then taking a, b, c common from C1, C2, and C3, we getΔ=−bcab+acac+abab+bc−acbc+abac+bcbc+ac−abNow, using C2→C2−C1 and C3→C3−C1, and then taking ( ab + bc + ca) common from C2 and C3, we getΔ=−bc11ab+bc−10ac+bc0−1×(ab+bc+ca)2Now, applying R2→R2+R1, we getΔ=−bc11ab01ac+bc0−1(ab+bc+ca)2Expanding along C2, we getΔ=(ab+bc+ca)2[ac+bc+ab]  =(ab+bc+ca)3  =(r/p)3=r3/p3Now given a, b, c are all positive, then A.M. ≥ G.M. ⇒ ab+bc+ac3≥(ab×bc×ac)1/3or   (ab+bc+ac)3≥27a2b2c2or   (ab+bc+ac)3≥27s2/p2If Δ=27, then ab + bc + ca = 3, and given that a2 + b2 + c2 = 3, from (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), we havea+b+c=±3⇒    a+b+c=3           ∵a, b, c∈R+⇒    3p+q=0Multiplying R1, R2, R3 by a, b, c, respectively, and then taking a, b, c common from C1, C2, and C3, we getΔ=−bcab+acac+abab+bc−acbc+abac+bcbc+ac−abNow, using C2→C2−C1 and C3→C3−C1, and then taking ( ab + bc + ca) common from C2 and C3, we getΔ=−bc11ab+bc−10ac+bc0−1×(ab+bc+ca)2Now, applying R2→R2+R1, we getΔ=−bc11ab01ac+bc0−1(ab+bc+ca)2Expanding along C2, we getΔ=(ab+bc+ca)2[ac+bc+ab]  =(ab+bc+ca)3  =(r/p)3=r3/p3Now given a, b, c are all positive, then A.M. ≥ G.M. ⇒ ab+bc+ac3≥(ab×bc×ac)1/3or   (ab+bc+ac)3≥27a2b2c2or   (ab+bc+ac)3≥27s2/p2If Δ=27, then ab + bc + ca = 3, and given that a2 + b2 + c2 = 3, from (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), we havea+b+c=±3⇒    a+b+c=3           ∵a, b, c∈R+⇒    3p+q=0