A cubic is more general than a quadratic, so clearly a quadratic is going to be a special case of a cubic.

We can demonstrate this by looking at the equation for a quadratic Bézier:

     (1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2

Now, let's increase the degree.

     = [ (1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2 ] * [ 1 - t + t ]

clearly this equation is equal to the above one because 1-t+t == 1 and multiplying by one does nothing.

     = [ (1 - t)^2 P_0 + 2(1 - t)t P_1 + t^2 P_2 ] * [ (1 - t) + t ]
     = (1 - t)^3 P_0 + t(1 - t)^2 P_0 + 2(1 - t)^2t P_1 + 2(1 - t)t^2 P_1 + (1 - t)t^2 P_2 + t^3 P_2

Group by powers of t:

     = (1 - t)^3 P_0 + t(1 - t)^2 [P_0 + 2*P_1] + (1 - t)t^2 [2P_1 + P_2] + t^3 P_2

Which is exactly a cubic Bezier curve.

Thus, all quadratic Beziers can represented cubic ones, but the reverse does not hold, thus quadratics are specializations of cubics.

QED.

This is known as degree elevation.

Oh, I see my mistake! Thank you very much!

I'm not very math savvy and I was trying to get this right.

Anyway, thank you very much for making this clear!

Cheers!