Multiple choice

Solve the system of the equations: $ax+by+cz=d,\quad { a }^{ 2 }x+{ b }^{ 2 }y+{ c }^{ 2 }z={ d }^{ 2 },\quad { a }^{ 3 }x+{ b }^{ 3 }y+{ c }^{ 3 }z={ d }^{ 3 }$.

  1. $x=\cfrac { d(d+b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a-d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $
  2. $x=\cfrac { d(d-b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a+d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $
  3. $x=\cfrac { d(d-b)(c-d) }{ a(a-b)(c-a) } ,\quad y=\cfrac { d(a-d)(d-c) }{ b(a-b)(b-c) } ,\quad z=\cfrac { d(b-d)(d-a) }{ d(b-c)(c-a) } $

  4. None of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
AI explanation

This system of equations represents finding the coefficients of a polynomial. We consider the polynomial P(t) = t cubed minus x times t squared minus y times t minus z, which has roots a, b, and c. Since P(a) = P(b) = P(c) = 0 based on the given equations, we can form the identity (t - a)(t - b)(t - c) = 0 and compare it to the structure formed by the given equations. Solving this using Lagrange interpolation or direct substitution leads directly to the formulas provided in option C.

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