Formula use:

a^3-b^3=\left(a-b\right)^3+3.a.b\left(a-b\right)

Explanation

According to the question,
\large\frac{x^2-1}{x}\normalsize=1
\therefore x-\large\frac{1}{x}\normalsize=1 _[lcm=x]
Here, given expression is :
\large\frac{x^6-1}{x^3}
\rightarrow x^3-\large\frac{1}{x^3} _[lcm=x³]
\rightarrow\left(x-\large\frac{1}{x}\right)\normalsize^3+3.x.\large\frac{1}{x}\normalsize\left(x-\large\frac{1}{x}\right)
Now, x-\large\frac{1}{x}\normalsize=1
\therefore\{\left(1\right)^3+3\times1\times1\}
\rightarrow\left(1+3\right)
\therefore 4
(proved)