Multiple choice

If $\alpha$ and $\beta$ be two zeros of the quadratic polynomial $ax^2+bx+c$, then $\dfrac {1}{\alpha^3}+\dfrac {1}{\beta^3}$ is equal to

  1. $\dfrac {3abc-b^3}{c^3}$
  2. $\dfrac {abc-b^3}{c^3}$
  3. $\dfrac {ac-b^3}{c^3}$
  4. $\dfrac {3abc-a^3}{c^3}$
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A Correct answer
Explanation

For ax^2+bx+c=0, alpha+beta = -b/a and alpha*beta = c/a. The expression 1/alpha^3 + 1/beta^3 = (alpha^3+beta^3)/(alpha*beta)^3. Using the identity alpha^3+beta^3 = (alpha+beta)^3 - 3*alpha*beta*(alpha+beta), the result simplifies to (3abc-b^3)/c^3.

AI explanation

By finding a common denominator, the expression becomes (beta^3 + alpha^3) / (alpha^3 beta^3), which equals ((alpha + beta)^3 - 3 alpha beta (alpha + beta)) divided by (alpha beta)^3. Using the coefficient relationships where alpha + beta = -b/a and alpha beta = c/a, the numerator becomes (-b/a)^3 - 3(c/a)(-b/a), which simplifies to (3abc - b^3) / a^3, and dividing by the denominator c^3 / a^3 yields (3abc - b^3) / c^3.

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