Multiple choice

If r,s,t are the roots of the equation ${ 8x }^{ 3 }+1001x+2008=0$ . the value of ${ \left( r+s \right) }^{ 3 }+{ \left( s+t \right) }^{ 3 }+{ \left( t+r \right) }^{ 3 }$ is

  1. $751$
  2. $752$
  3. $753$
  4. $754$
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C Correct answer
Explanation

Given r, s, t are roots of 8x^3 + 1001x + 2008 = 0. By Vieta's formulas, r+s+t = 0. Thus r+s = -t, s+t = -r, t+r = -s. The expression becomes (-t)^3 + (-r)^3 + (-s)^3 = -(r^3 + s^3 + t^3). Using the identity for roots of cubic equations, the result is 753.

AI explanation

From the cubic equation 8x^3 + 1001x + 2008 = 0, the sum of the roots r + s + t = 0 because there is no x^2 term. Expanding the target expression (r + s)^3 + (s + t)^3 + (t + r)^3 and applying the identity a^3 + b^3 + c^3 = 3abc when a + b + c = 0, we substitute a = r + s, b = s + t, and c = t + r. This simplifies the expression to 3(r + s)(s + t)(t + r). Since r + s + t = 0, we can write this as 3(-t)(-r)(-s), which equals -3rst. The product of the roots for the cubic equation is given by -2008/8, so rst = -251. Substituting this gives -3(-251) = 753.

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