Cubing a Binomial

What will we obtain if we cube a general binomial?

\({\left( {x + y} \right)^3} = ?\)

We have

\[\begin{array}{l}{\left( {x + y} \right)^3} = \left( {x + y} \right) \times {\left( {x + y} \right)^2}\\\qquad\quad\;\;\;= \left( {x + y} \right)\left( {{x^2} + 2xy + {y^2}} \right)\end{array}\]

Now, we multiply these two brackets term-by-term:

\[\begin{array}{l}{\left( {x + y} \right)^3} = \left( {x + y} \right)\left( {{x^2} + 2xy + {y^2}} \right)\\ \qquad\qquad= \left\{ \begin{array}{l}x \times \left( {{x^2} + 2xy + {y^2}} \right)\\ \qquad \qquad + \\y \times \left( {{x^2} + 2xy + {y^2}} \right)\end{array} \right.\\  \qquad\qquad= \left\{ \begin{array}{l}{x^3} + 2{x^2}y + x{y^2} + \\{x^2}y + 2x{y^2} + {y^3}\end{array} \right.\end{array}\]

Thus,

\({\left( {x + y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3}\)

This is an identity – it holds true for every value of x and y. If we replace \(y \to  - y,\) we have:

\({\left( {x - y} \right)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3}\)

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