# Squaring a Trinomial

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What will we get if we expand a general trinomial?

$${\left( {x + y + z} \right)^2} = ?$$

We have

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

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

$\begin{array}{l}{\left( {x + y + z} \right)^2} = \left( {x + y + z} \right)\left( {x + y + z} \right)\\ \qquad\qquad\qquad\! \! = \left\{ \begin{array}{l}x \times \left( {x + y + z} \right)\\ \qquad\qquad+ \\y \times \left( {x + y + z} \right)\\ \qquad\qquad+ \\z \times \left( {x + y + z} \right)\end{array} \right.\\ \qquad\qquad\qquad\! \! = \left\{ \begin{array}{l}{x^2} + xy + xz + \\xy + {y^2} + yz + \\xz + yz + {z^2}\end{array} \right.\end{array}$

Thus,

$${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + xz} \right)$$

This is again an identity, an equality which holds true for every value of x, y and z.