Factorize a³ + (b − a)³ − b³
The factorized form of \[ a^3+(b-a)^3-b^3 \] is
Solution
Let
\[ x=a,\quad y=(b-a),\quad z=-b \]
Then,
\[ x+y+z = a+(b-a)-b \]
\[ x+y+z=0 \]
We know that if \(x+y+z=0\), then
\[ x^3+y^3+z^3=3xyz \]
Therefore,
\[ a^3+(b-a)^3-b^3 = 3(a)(b-a)(-b) \]
\[ = -3ab(b-a) \]
\[ = 3ab(a-b) \]
Therefore,
\[ \boxed{3ab(a-b)} \]