Ireland_2017

With $s=t-1$ we have $s^3=t^3-3t^2+3t-1$, and the first equation can be written as $x=s^3+1$. From the second equation we get $27(s^3{)}^4=(x^2-1{)}^3$, hence, by taking cube roots, $3s^4=x^2-1$, i.e. $x^2=3s^4+1$. Comparing with the square of $x=s^3+1$ we obtain $$3s^4+1=s^6+2s^3+1,\text{ or}$$ $$s^6-3s^4+2s^3=0,\text{i.e.}$$ $$s^3(s-1{)}^2(s+2)=0.$$ So, $s=-2,0,1$, i.e., $t=s+1=-1,1,2$ and $x=s^3+1=-7,1,2$. The three solutions $(t,x)$ are $(-1,-7)$, $(1,1)$ and $(2,2)$.