Rewrite \({x}^{3}-1\) in the form \({a}^{3}-{b}^{3}\), where \(a=x\) and \(b=1\).
\[\int {x}^{3}-{1}^{3} \, {dx}^{2}dx\]
Use Difference of Cubes: \({a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})\).
\[\int (x-1)({x}^{2}+(x)(1)+{1}^{2}) \, {dx}^{2}dx\]
Simplify \({1}^{2}\) to \(1\).
\[\int (x-1)({x}^{2}+x\times 1+1) \, {dx}^{2}dx\]
Simplify \(x\times 1\) to \(x\).
\[\int (x-1)({x}^{2}+x+1) \, {dx}^{2}dx\]
Expand.
\[\int {x}^{3}-1 \, {dx}^{2}dx\]
Use Power Rule: \(\int {x}^{n} \, dx=\frac{{x}^{n+1}}{n+1}+C\).
\[{x}^{3}d{(\frac{{x}^{3}}{4}-1)}^{2}\]
x^3*d*(x^3/4-1)^2
