Find the interpolating polynomial of degree $3$ that interpolates $f(x) = x^3$ at the nodes $x_0=0, x_1 = 1, x_2=2, x_3 = 3$.
Here are my workings below
The basic Lagrange polynomials are:
$$L_0(x) = \frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)}$$
$$L_1(x) = \frac{(x-0)(x-2)(x-3)}{(1-0)(1-2)(1-3)}$$
$$L_2(x) = \frac{(x-0)(x-1)(x-3)}{(2-0)(2-1)(2-3)}$$
$$L_3(x) = \frac{(x-0)(x-1)(x-2)}{(3-0)(3-1)(3-2)}$$
Then the interpolating polynomial is:
$$P(x) = L_0(x)+(1)^3L_1(x)+(2)^3L_2(x)+(3)^3L_3(x)$$
Am I allowed to find the interpolating polynomial using these basic lagrange polynomials? and is my $P(x)$ correct? I wasn't sure if the first term should be $L_0(x)$?