f the solutions of the equation x2 + px + q = 0 are the cubes of the solutions of the equation x2 + mx + 2 = 0, then
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• p – q = m3
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• p = m3q
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• p + q = m3
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• p = m3 + 6m
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• p = m3 – 6m

• let roots of eqn x2+mx+2=0 are a and b.
  a+b=-m, ab=2
  and roots of eqn x2+px+q=0 are a3 and b3.
  a3+b3=-p, a3*b3=q
  now,we know,
  a3+b3=(a+b)(a2+b2-ab);
  a3+b3=(a+b)[(a+b)2-3ab];
  putting the values of all by above,
  -p=(-m)(m2-6)
  p=m3-6m
  hence ans=(e)
• 12 years agoHelpfull: Yes(34) No(0)
• let roots of equation x^2+mx+y=0 are a and b.
  a+b=-m
  ab=2
  (a+b)^3=a^3+b^3+3ab(a+b)
  (-m)^3=a^3+b^3+(3*2*(-m))
  -m^3=a^3+b^3-6m
  a^3+b^3=-m^3+6m
  a^3+b^3 are the sum of roots of x^2+px+q
  so a^3+b^3=-p
  -m^3+6m=-p
  p=m^3-6m
• 10 years agoHelpfull: Yes(4) No(0)
• ans is p=m^3-6m
• 12 years agoHelpfull: Yes(2) No(0)