$={\left(3x^2\right)}_{x=t}=3t^2$

So equation tangent at $P\left(t,t^3\right):$

$y-t^3=3t^2\left(x-t\right)$
for point of intersection with $y=x^3$

$x^3-t^3=3t^{2}x-3t^3$

$\Rightarrow \left(x-t\right)\left(x^2+xt+t^2\right)=3t^2\left(x-t\right)$

for $x\ne t$

$x^2+xt+t^2=3t^2$

$\Rightarrow x^2+xt-2t^2=0\Rightarrow \left(x-t\right)\left(x+2t\right)=0$

So for $Q:x=-2t,Q\left(-2t,-8t^3\right)$
ordinate of required point $:\frac{2t^3-8t^3}{2+1}=-2t^3$