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    গণিত

    Topic

    Algebra

    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.

    ক)
    p1 = −a; p2 = a^2 − 2b; p3 = −a^3 + 3ab − 3c
    খ)
    p1 = a; p2 = a^2 + 2b; p3 = a^3 + 3ab + 3c
    গ)
    p1 = −a; p2 = a^2 + 2b; p3 = −a^3 − 3ab + 3c
    ঘ)
    p1 = −b; p2 = b^2 − 2c; p3 = −b^3 + 3bc − 3a

    Explanation

    Let the monic cubic be f(x)=x3+ax2+bx+cf(x)=x^{3}+ax^{2}+bx+c with roots r1,r2,r3r_{1},r_{2},r_{3}. By Viète's relations e1=r1+r2+r3=−a,e2=r1r2+r2r3+r3r1=b,e3=r1r2r3=−c. 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 pk=r1k+r2k+r3kp_{k}=r_{1}^{k}+r_{2}^{k}+r_{3}^{k}. Then 1) p1=e1=−a.p_{1}=e_{1}=-a. 2) For p2p_{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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