If ${\mathrm{x}}^2+\mathrm{x}+1$ is a factor of ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$ the real roots of ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}=0$ is

We know that, x ²+ x + 1 is a factor of ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$
Hence, roots of x ²+ x + 1 are also roots of 
${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$. Since, $\mathrm{\omega } \text{and }{\mathrm{\omega }}^2$

$\left(\text{where}\mathrm{\omega }=-\frac{1}{2}+\frac{3\mathrm{i}}{2}\right)$ are two complex roots  x ²+ x + 1

Therefore, $\mathrm{\omega } \text{and }{\mathrm{\omega }}^2$ are two complex roots of ${\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$

We know that, a cubic equation has atleast one real root. Let real root be $\alpha$. Then,

$\mathrm{\alpha }\cdot \mathrm{\omega }\cdot {\mathrm{\omega }}^2=-\frac{\mathrm{d}}{\mathrm{a}}\Rightarrow \mathrm{\alpha }=-\frac{\mathrm{d}}{\mathrm{a}}$