Multiple choice

If $\alpha,\beta,\gamma$ are the roots of the equation $\mathrm{x}^{3}+4\mathrm{x}+1=0$, then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}=$

  1. $2$
  2. $3$
  3. $4$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given alpha + beta + gamma = 0, alpha*beta + beta*gamma + gamma*alpha = 4, and alpha*beta*gamma = -1. We need 1/(alpha+beta) + 1/(beta+gamma) + 1/(gamma+alpha). Since alpha+beta = -gamma, the expression is -1/gamma - 1/alpha - 1/beta = -(alpha*beta + beta*gamma + gamma*alpha) / (alpha*beta*gamma) = -4 / -1 = 4.