If you're not very convincing, convince yourself that the polynomial is not cubic. When you fail, you will be convinced that it is cubic. QED

It's not too bad, once you consider that everything in each term is a constant value, except for "x" itself.

So the numerator of each term is the product of three linear factors, like (x-4)(x-2)(x-6), which should produce a cubic, like x³ - 12x² + 44x - 48. Then the denominator of each term is a pure constant, so it would be like taking that cubic and dividing it by 4, getting 0.25x³ - 3x² + 11x - 12. Then the yₙ terms are also constants, so no different than doing something like multiplying by 2, getting you something like 0.5x³ - 6x² + 22x - 24, if I take that example a bit too far. And at that point it's just the sum of four cubics, which will be cubic as long as the x³ terms don't perfectly cancel out - which I believe would only happen if the four pairs of points used to make the function were all perfectly laying on the same line or parabola.

The construction's also pretty clever: OP said the point was to fit the function to the four points (x₁, y₁), (x₂, y₂), (x₃, y₃), (x₄, y₄). Let's say we set x = x₂, then. Because (x-x₂) appears in the numerator of every term but the second term, every term but the second term will have a 0 in its numerator and cancel out - so we only need to consider the second term. Its numerator is then (x₂-x₁)(x₂-x₃)(x₂-x₄) - exactly the same as its denominator. So they both cancel out, leaving only y₂ - meaning we get P₃(x₂) = y₂, as desired.

Sure I can convince myself, but how am I going to get Yi to agree? He's known to be rather stubborn.

I'm not sure sure what the stated premise is supposed to even mean.

Would 'I looked it through and as a result of that, I am convinced' be an acceptable answer?

If you have a series of (nx + m) terms multiplied together (where each n and m can be different), just count the number of times x comes up (let's call that z), and the order of the function will be x^z. For z = 3, that's cubic. None of the other terms matter for the expansion. If there's an x^3 and no higher powers of x, it will always be cubic and the coefficients of any of the terms are irrelevant.

If you're studying computer science, you can think of it like big O notation.

And for the other part, one trick for questions like that is to look at what happens when you substitute an equality they are talking about, specifically looking for terms that result in 0s or 1s and can be cancelled out of the equation. If you're not given actual values for any coefficients, there's not much else interesting you can do to a polynomial like that, other than maybe take the derivative or integrate.