Question : If $\alpha$ and $\beta$ are the roots of equation $x^{2}-x+1=0$, then which equation will have roots $\alpha ^{3}$ and $\beta ^{3}?$
Option 1: $x^{2}+2x+1=0$
Option 2: $x^{2}-2x-1=0$
Option 3: $x^{2}+3x-1=0$
Option 4: $x^{2}-3x+1=0$
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Correct Answer: $x^{2}+2x+1=0$
Solution : $\alpha$ and $\beta$ are roots of the equation $x^{2}-x+1=0$ ⇒ $\alpha+\beta=1$ and $\alpha\beta=1$ Now we need to find the equation whose roots are $\alpha ^{3}$ and $\beta ^{3}$. ⇒ $(\alpha^{3}+\beta^{3})=(\alpha+\beta)^{3}-3\alpha\beta\times(\alpha+\beta)$ ⇒ $(\alpha^{3}+\beta^{3})=(1)^{3}-3\times 1 \times (1)$ ⇒ Sum of roots = $(\alpha^{3}+\beta^{3})=-2$ Also, Product of roots = $(\alpha^{3}\beta^{3})=(\alpha\beta)^{3} =1$ So, the required equation is $x^{2}+2x+1=0$ Hence, the correct answer is $x^{2}+2x+1=0$.
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