Factorization of (x + y)³ − (x − y)³
\[ (x+y)^3-(x-y)^3 \] can be factorized as
(a) \(2y(3x^2+y^2)\)
(b) \(2x(3x^2+y^2)\)
(c) \(2y(3y^2+x^2)\)
(d) \(2x(x^2+3y^2)\)
Solution
\[ (x+y)^3=x^3+3x^2y+3xy^2+y^3 \]
\[ (x-y)^3=x^3-3x^2y+3xy^2-y^3 \]
\[ (x+y)^3-(x-y)^3 \]
\[ =x^3+3x^2y+3xy^2+y^3 -x^3+3x^2y-3xy^2+y^3 \]
\[ =6x^2y+2y^3 \]
\[ =2y(3x^2+y^2) \]
Therefore,
\[ \boxed{(a)\ 2y(3x^2+y^2)} \]