Find alpha^4+beta^4 if alpha, beta are the ro | vedclass

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(C) The given equation is $x^2+x+1=0$. Since the roots are $\alpha$ and $\beta$,we know that $\alpha$ and $\beta$ are the non-real cube roots of unity

Vedclass · Accepted Answer

$-1$

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(C) The given equation is $x^2+x+1=0$. Since the roots are $\alpha$ and $\beta$,we know that $\alpha$ and $\beta$ are the non-real cube roots of unity,i.e.,$\omega$ and $\omega^2$. Thus,$\alpha^3 = 1$ and $\beta^3 = 1$. Also,from the equation,$\alpha + \beta = -1$ and $\alpha \beta = 1$. We need to find $\alpha^4 + \beta^4$. Since $\alpha^3 = 1$,$\alpha^4 = \alpha^3 \cdot \alpha = 1 \cdot \alpha = \alpha$. Similarly,$\beta^4 = \beta^3 \cdot \beta = 1 \cdot \beta = \beta$. Therefore,$\alpha^4 + \beta^4 = \alpha + \beta$. Substituting the value of $\alpha + \beta$,we get $\alpha^4 + \beta^4 = -1$.

Find $\alpha^4+\beta^4$ if $\alpha, \beta$ are the roots of the equation $x^2+x+1=0$.

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