If α and β are the roots of equation x 2– x + 1 = 0, then which equation will have roots α 3and β 3?
x2 + 2x + 1 = 0
For \(x^2-x+1=0\), by Vieta's: \(\alpha+\beta=1\), \(\alpha\beta=1\).
Step 1 — Sum of cubes:
\[\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=1-3=-2\]
Step 2 — Product of cubes: \(\alpha^3\beta^3=(\alpha\beta)^3=1\).
Step 3 — Form the equation: \(x^2-(\text{sum})x+(\text{product})=0\):
\[x^2-(-2)x+1=0 \implies x^2+2x+1=0\]
Therefore the required equation is x² + 2x + 1 = 0.
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