If α, β are roots of the equation x

Solution:

Given : $\alpha$ and $\beta$ are roots of $x^{2}+x+1=0$

i. e. $\alpha^{2}+\alpha+1=0$ and $\beta^{2}+\beta+1=0$

Since $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$

$\Rightarrow \alpha^{3}-1=(\alpha-1)\left(\alpha^{2}+\alpha+1\right)$

and $\beta^{3}-1=(\beta-1)\left(\beta^{2}+\beta+1\right)$

$\Rightarrow \alpha^{3}-1=0$ and $\beta^{3}-1=0$

i.e. $\alpha^{3}=1$ and $\beta^{3}=1$

Now, $\alpha^{19}+\beta^{7}=\left(\alpha^{3}\right)^{6} \cdot \alpha+\left(\beta^{3}\right)^{2} \beta$

$=1 \cdot \alpha+1 \cdot \beta \quad\left(\because \alpha^{3}=1\right.$ and $\left.\beta^{3}=1\right)$

$=\alpha+\beta=-1$

$\left(\because \alpha\right.$ and $\beta$ are roots of $\left.x^{2}+x+1\right)$

i. e. $\alpha^{19}+\beta^{7}=-1$

and $\alpha^{19} \beta^{7}=\left(\alpha^{3}\right)^{6} \alpha\left(\beta^{3}\right)^{2} \beta$

$=\alpha \beta=1 \quad\left(\because \alpha\right.$ and $\beta$ are roots of $\left.x^{2}+x+1\right)$

Hence $\alpha^{19}$ and $\beta^{7}$ satisfies quadratic equation $x^{2}-(-1) x+1=0$

i. e. $x^{2}+x+1=0$