Important Polynomial Identity

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Now, we discuss a really important identity. We start with considering the following product:

\[P = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)\]

There are three terms in the first bracket and six terms in the second product; lets multiply them term-by-term:

\[\begin{array}{l}P = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)\\\,\,\,\,\, = \left\{ \begin{array}{l}x \times \left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right) + \\y \times \left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right) + \\z \times \left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)\end{array} \right.\\\,\,\,\,\, = \left\{ \begin{array}{l}{x^3} + \overbrace {x{y^2}}^{\rm{I}} + \overbrace {x{z^2}}^{{\rm{II}}}\overbrace { - {x^2}y}^{{\rm{III}}} - xyz\overbrace { - {x^2}z}^{{\rm{IV}}} + \\\overbrace {{x^2}y}^{{\rm{III}}} + {y^3} + \overbrace {y{z^2}}^{\rm{V}}\overbrace { - x{y^2}}^{\rm{I}}\overbrace { - {y^2}z}^{{\rm{VI}}} - xyz + \\\overbrace {{x^2}z}^{{\rm{IV}}} + \overbrace {{y^2}z}^{{\rm{VI}}} + {z^3} - xyz\overbrace { - y{z^2}}^{\rm{V}}\overbrace { - x{z^2}}^{{\rm{II}}}\end{array} \right.\end{array}\]

Some of the terms cancel out, as shown above (terms which cancel out are marked with the same Roman numerals), and we are left with:

\[\begin{array}{l}P = {x^3} - xyz + {y^3} - xyz + {z^3} - xyz\\\;\,\,\,\,\, = {x^3} + {y^3} + {z^3} - 3xyz\end{array}\]

Thus, the identity is:

\(\begin{array}{l}\left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right) = \\{x^3} + {y^3} + {z^3} - 3xyz\end{array}\)

Example 1: Expand the following:

\({\left( {1 + x + {x^2}} \right)^2}\)

Solution: Using the relevant identity, we have:

\[\begin{array}{l}{\left( {1 + x + {x^2}} \right)^2}\\ = \left\{ \begin{array}{l}{1^2} + {x^2} + {\left( {{x^2}} \right)^2} + \\2\left( 1 \right)\left( x \right) + 2\left( x \right)\left( {{x^2}} \right) + 2\left( 1 \right)\left( {{x^2}} \right)\end{array} \right.\\ = 1 + {x^2} + {x^4} + 2x + 2{x^3} + 2{x^2}\\ = 1 + 2x + 3{x^2} + 2{x^3} + {x^4}\end{array}\]

Example 2: Expand the following:

\({\left( {1 + x} \right)^3}{\left( {1 - x} \right)^3}\)

Solution: Instead of cubing the two binomials separately, we note the following:

\[{\left( {1 + x} \right)^3}{\left( {1 - x} \right)^3} = {\left( {1 - {x^2}} \right)^3}\]

The expression on the right is much easier to cube:

\[\begin{array}{l}{\left( {1 - {x^2}} \right)^3}\\ = {\left( 1 \right)^3} - 3{\left( 1 \right)^2}\left( {{x^2}} \right) + 3\left( 1 \right){\left( {{x^2}} \right)^2} - {\left( {{x^2}} \right)^3}\\ = 1 - 3{x^2} + 3{x^4} - {x^6}\end{array}\]

Example 3: Factorize \({x^3} - {y^3} - x + y\).

Solution: We have:

\[\begin{array}{l}{x^3} - {y^3} - x + y\\ = \left( {{x^3} - {y^3}} \right) - \left( {x - y} \right)\\ = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) - \left( {x - y} \right)\\ = \left( {x - y} \right)\left( {{x^2} + xy + {y^2} - 1} \right)\end{array}\]

Example 4: Factorize \({a^6} - 7{a^3} - 8\).

Solution: Letting \({a^3} = x\) reduces this to a quadratic which can be easily factorized:

\[\begin{array}{l}{a^6} - 7{a^3} - 8 = {x^2} - 7x - 8\\ \;\;\qquad\qquad = \left( {x - 8} \right)\left( {x + 1} \right)\\ \;\;\qquad\qquad= \left( {{a^3} - 8} \right)\left( {{a^3} + 1} \right)\end{array}\]

Both these terms can now be factorized separately:

\[\begin{array}{l}\left( {{a^3} - 8} \right)\left( {{a^3} + 1} \right)\\ = \left\{ {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)} \right\}\left\{ {\left( {a + 1} \right)\left( {{a^2} - a + 1} \right)} \right\}\\ \Rightarrow \,\,\,{a^6} - 7{a^3} - 8\\\,\,\,\,\,\,\, = \left( {a - 2} \right)\left( {a + 1} \right)\left( {{a^2} + 2a + 4} \right)\left( {{a^2} - a + 1} \right)\end{array}\]

Example 5: Factorize \({x^3} + 1 - {y^3} + 3xy\).

Solution: We have:

\[\begin{array}{l}{x^3} + 1 - {y^3} + 3xy\\ = {\left( x \right)^3} + {\left( { - y} \right)^3} + {\left( 1 \right)^3} - 3\left( a \right)\left( { - y} \right)\left( 1 \right)\end{array}\]

Now, we make use of the relevant identity we discussed a while back:

\[\begin{array}{l}{x^3} + 1 - {y^3} + 3xy\\ = \left( {x - y + 1} \right)\left( {{x^2} + {y^2} + xy - x + y + 1} \right)\end{array}\]

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