You assumption is incorrect. At each instant the load power is shared amongst the three phases.
Imagine for simplicity the voltages and currents are in phase. It works out even if they are not, but is more difficult to see.
Consider phase voltages:
$$
V_a = V\cos(\omega t), V_b = V\cos(\omega t - 120°), V_c = V\cos(\omega t + 120°)
$$
Then at \$t=0\$, power in phase A is zero, but the other two are non-zero. A little while later all three phases will be non-zero. After a third of a cycle, power in phase B will reach zero while the other two are non-zero.
So it can be seen that at least two of the individual instantaneous phase powers are usually non-zero.
To see that the total power is nonetheless constant, use the cosine multiplication trig identity \$\cos\alpha \cos\beta = \frac{1}{2}(\cos(\alpha + \beta)+\cos(\alpha - \beta))\$ to multiply voltage and current:
$$
P_a = V\cos(\omega t) \cdot I\cos(\omega t) = \frac{VI}{2}[\cos(2\omega t)+\cos(0°)]
$$
$$
P_b = V\cos(\omega t- 120°) \cdot I\cos(\omega t- 120°) = \frac{VI}{2}[\cos(2\omega t+120°)+\cos(0°)]
$$
$$
P_c = V\cos(\omega t+ 120°) \cdot I\cos(\omega t+ 120°) = \frac{VI}{2}[\cos(2\omega t-120°)+\cos(0°)]
$$
Since \$\cos(0°) = 1\$ the sum of powers is:
$$
P_a+P_b+P_c = \frac{VI}{2}[\cos(2\omega t)+\cos(2\omega t+120°)+\cos(2\omega t-120°)]+\frac{3VI}{2}
$$
Clearly \$\frac{3VI}{2}\$ is constant, so we need only concentrate on the other part. Applying the cosine multiplication trig identity in reverse this time, we have:
$$
\cos(2\omega t)+\cos(2\omega t+120°)+\cos(2\omega t-120°) = \cos(2\omega t)+2\cos(2\omega t)\cos(120°)
$$
Since \$\cos(120°) = -\frac{1}{2}\$ this reduces to zero, leaving just the constant part. Thus, the instantaneous combined power of the three phases does not depend on \$t\$. In other words, it "is the same at any instant".
A graphical representation of these equations is shown in this graph.

\$P_a\$ is red, \$P_b\$ is yellow, \$P_c\$ is blue and the sum is orange.