Answer: To show that the curve passes through a point, we need to find a value of t that makes the parametric equations satisfy the coordinates of the point.
Let's first check if the curve passes through the point (1, 4, 0):
x = t^2, so when x = 1, we have t = ±1.
y = 1 - 3t, so when t = 1, we have y = -2.
z = 1 + t^3, so when t = 1, we have z = 2.
Therefore, the curve passes through the point (1, 4, 0).
Next, let's check if the curve passes through the point (9, -8, 28):
x = t^2, so when x = 9, we have t = ±3.
y = 1 - 3t, so when t = -3, we have y = 10.
z = 1 + t^3, so when t = 3, we have z = 28.
Therefore, the curve passes through the point (9, -8, 28).
Finally, let's check if the curve passes through the point (4, 7, -6):
x = t^2, so when x = 4, we have t = ±2.
y = 1 - 3t, so when t = 2, we have y = -5.
z = 1 + t^3, so when t = 2, we have z = 9.
Therefore, the curve does not pass through the point (4, 7, -6).
Hence, we have shown that the curve passes through the points (1, 4, 0) and (9, -8, 28) but not through the point (4, 7, -6).