Discussion
The expression
\[ \begin{aligned} &(x - y)(x^{2} + xy + y^{2}) + (x + y)(x^{2} - xy + y^{2}) \\ &- (x + y)(x^{2} - y^{2}) \end{aligned} \]
is equal to
Answer: C
The expression equals \(x^{3} + y^{3} + xy(y - x)\).
Use the identities \((x - y)(x^{2} + xy + y^{2}) = x^{3} - y^{3}\) and \((x + y)(x^{2} - xy + y^{2}) = x^{3} + y^{3}\):
\[ \begin{aligned} &(x^{3} - y^{3}) + (x^{3} + y^{3}) - (x + y)(x^{2} - y^{2}) \\ &= 2x^{3} - (x^{3} + x^{2}y - xy^{2} - y^{3}) \\ &= x^{3} + y^{3} - x^{2}y + xy^{2} \\ &= x^{3} + y^{3} + xy(y - x) \end{aligned} \]
The option with \(xy(x - y)\) has the sign of the last term reversed. Check with x = 2, y = 1: the expression is 7 + 9 − 9 = 7, and \(8 + 1 + 2(1 - 2) = 7\).
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