If $x^6+x^5+x^4+x^3+x^2+x+1=0$ then what is the value of $x^{42}+x^{84}$.

$x^6+x^5+x^4+x^3+x^2+x+1=0$   ...(i)

$\Rightarrow$When multiply by x, we get

$\Rightarrow x^7+x^6+x^5+x^4+x^3+x^2+x=0$

$\Rightarrow x^7-1=0$

$\Rightarrow x^7=1$

The value of $x^{42}+x^{84}={\left(x^7\right)}^6+{\left(x^7\right)}^{12}={\left(1\right)}^6+{\left(1\right)}^{12}=2.$