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: Use Vieta's relations together with Newton's identity for the sum of cubes instead of expanding individual roots.

Step 1: For \(x^4+x^3-x-1=0\), Vieta's formulas give \(\alpha+\beta+\gamma+\delta=-1\), \(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=0\) and \(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=1\).

Step 2: Using \(p_3=(\alpha+\beta+\gamma+\delta)^3-3(\alpha+\beta+\gamma+\delta)(\sum\alpha\beta)+3(\sum\alpha\beta\gamma)\), we get \(p_3=(-1)^3-3(-1)(0)+3(1)=2\).

Step 3: Hence, \(\boxed{\alpha^3+\beta^3+\gamma^3+\delta^3=2}\).