Q1Advanced Engineering Mathematics I

Given the expression: $f(x)=(ax-1)(b{x}^2-1)(c{x}^3-1)$

Expanding the expression gives a polynomial of degree $1+2+3=6$. The leading term is $(a)(b)(c)x^6=abc{x}^6$.

We know that the $n$-th forward difference of a polynomial of degree $n$ is a constant, and it is given by: ${\Delta }^n(k{x}^n)=k\cdot n!\cdot h^n$.

Assuming the interval of differencing $h=1$: ${\Delta }^6(abc{x}^6)=abc\cdot 6!=abc\cdot 720=720abc$.

Since the 6th differences of all lower degree terms are zero, the answer is $720abc$.