Q.If $h = 1$ then prove that $(E^{-1}\Delta)x^3 = 3x^2 - 3x + 1$.

$(E^{-1}\Delta)x^3 = x^3 - (x-1)^3 = 3x^2 - 3x + 1$.

Step 1 — simplify the operator. Using $\Delta = E - 1$,

$E^{-1}\Delta = E^{-1}(E - 1) = 1 - E^{-1} = \nabla,$

the backward difference operator.

Step 2 — apply it to $x^3$ with $h = 1$. Since $E^{-1}f(x) = f(x - h) = f(x-1)$,

$(E^{-1}\Delta)x^3 = (1 - E^{-1})x^3 = x^3 - (x-1)^3.$

Step 3 — expand $(x-1)^3$.

$(x-1)^3 = x^3 - 3x^2 + 3x - 1.$

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