Multiple choice

If $\alpha , \beta $ and $\gamma $ are the roots of the equation ${ x }^{ 3 }-8x+8=0$, then $\displaystyle\sum { { \alpha }^{ 2 } } $ and $\displaystyle\sum { \dfrac { 1 }{ \alpha \beta } } $ are respectively

  1. $0$ and $-16$
  2. $16$ and $8$
  3. $-16$ and $0$
  4. $16$ and $0$
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D Correct answer
Explanation

For x^3 - 8x + 8 = 0, the coefficients are a=1, b=0, c=-8, d=8. Sum of roots (alpha+beta+gamma) = -b/a = 0. Sum of roots taken two at a time (alpha*beta + beta*gamma + gamma*alpha) = c/a = -8. Product of roots (alpha*beta*gamma) = -d/a = -8. Sum of squares = (sum of roots)^2 - 2*(sum of roots taken two at a time) = 0^2 - 2*(-8) = 16. Sum of 1/(alpha*beta) = (alpha+beta+gamma)/(alpha*beta*gamma) = 0 / -8 = 0.

AI explanation

Using Vieta's formulas for x^3 - 8x + 8 = 0, the sum of roots (alpha + beta + gamma) is 0, the sum of product of roots two at a time is -8, and the product of roots is -8. The sum of squares of the roots is calculated as (alpha + beta + gamma)^2 - 2(alpha*beta + beta*gamma + gamma*alpha), which gives 0^2 - 2(-8) = 16. The sum of 1/(alpha*beta) evaluates to (alpha + beta + gamma)/(alpha*beta*gamma), resulting in 0 / -8 = 0.

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