Question

If \(\alpha,\beta,\gamma,\delta\) are the roots of the equation \[ x^4+x^3-x-1=0, \] then \[ \alpha^3+\beta^3+\gamma^3+\delta^3= \]

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-2\) \bigskip

Hint:

For questions involving sums of powers of roots, first find the elementary symmetric sums using Vieta's formulas. Then use standard identities or Newton's sums to evaluate higher powers efficiently.

Answer 1

The Correct Option is C

Concept: If \(\alpha,\beta,\gamma,\delta\) are the roots of \[ x^4+a_1x^3+a_2x^2+a_3x+a_4=0, \] then by Vieta's formulas, \[ \alpha+\beta+\gamma+\delta=-a_1, \] \[ \alpha\beta+\beta\gamma+\gamma\delta+\delta\alpha+\alpha\gamma+\beta\delta=a_2. \] Also, \[ \alpha^3+\beta^3+\gamma^3+\delta^3 = (\alpha+\beta+\gamma+\delta)^3 -3(\alpha+\beta+\gamma+\delta) (\alpha\beta+\beta\gamma+\gamma\delta+\delta\alpha+\alpha\gamma+\beta\delta) +3(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta). \]

Step 1: Find the symmetric sums using Vieta's formulas. Given \[ x^4+x^3+0x^2-x-1=0. \] Comparing with \[ x^4+a_1x^3+a_2x^2+a_3x+a_4=0, \] we get \[ a_1=1,\qquad a_2=0,\qquad a_3=-1. \] Therefore, \[ \alpha+\beta+\gamma+\delta=-1, \] \[ \alpha\beta+\beta\gamma+\gamma\delta+\delta\alpha+\alpha\gamma+\beta\delta=0, \] and \[ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=1. \]

Step 2: Apply the formula for the sum of cubes. Substituting the values, \[ \alpha^3+\beta^3+\gamma^3+\delta^3 = (-1)^3 -3(-1)(0) +3(1). \] \[ =-1+0+3. \] \[ =2. \]

Step 3: Write the final answer. \[ \boxed{2} \]