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If $p, q, r$ are positive integers and $\omega$ is the cube root of unity and $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3 \mathrm{p}}+\mathrm{x}^{3 \mathrm{q}}+1+\mathrm{x}^{3 \mathrm{r}+2}$, then what is $\mathrm{f}(\omega)$ equal to?
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Since $\omega$ is a cube root of unity $\therefore \omega^{3}=1$ and $1+\omega+\omega^{2}=0$
Let $\mathrm{f}(\mathrm{x})=\left(x^{3}\right) p+\left(x^{3}\right)^{\mathrm{q}} \cdot x+\left(x^{3}\right)^{r} \cdot x^{2}$
Now, put $x=\omega$ $\mathrm{f}(\omega)=\left(\omega^{3}\right)^{p}+\left(\omega^{3}\right)^{q} . \omega+\left(\omega^{3}\right)^{r} \cdot \omega^{2}$
$=1^{p}+1^{q} \cdot \omega+1^{r} . \omega^{2}=1+\omega+\omega^{2}=0$
Let $\mathrm{f}(\mathrm{x})=\left(x^{3}\right) p+\left(x^{3}\right)^{\mathrm{q}} \cdot x+\left(x^{3}\right)^{r} \cdot x^{2}$
Now, put $x=\omega$ $\mathrm{f}(\omega)=\left(\omega^{3}\right)^{p}+\left(\omega^{3}\right)^{q} . \omega+\left(\omega^{3}\right)^{r} \cdot \omega^{2}$
$=1^{p}+1^{q} \cdot \omega+1^{r} . \omega^{2}=1+\omega+\omega^{2}=0$
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