 QUESTION

# Factorise the given expression $27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz$

Hint: Use the identity ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz \right)$

Express the given expression in the form of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz$ and use the above-mentioned property.

Before solving the question, we need to know the meaning of factorisation. Consider two algebraic expressions ${{a}^{2}}-{{b}^{2}}$ and $\left( a+b \right)\left( a-b \right)$. Let us simplify the latter expression. Applying distributive property we get $\left( a+b \right)\left( a-b \right)=\left( a+b \right)a-\left( a+b \right)b$
Applying distributive property again we get
$\left( a+b \right)\left( a-b \right)={{a}^{2}}+ab-\left[ ab+{{b}^{2}} \right]$
Simplifying, we get
\begin{align} & \left( a+b \right)\left( a-b \right)={{a}^{2}}+ab-ab-{{b}^{2}} \\ & ={{a}^{2}}-{{b}^{2}} \\ \end{align}
Hence the two expressions are equal.
The expression $\left( a+b \right)\left( a-b \right)$is said to be factorised form of ${{a}^{2}}-{{b}^{2}}$. When factorising an expression, we make use of algebraic identities like ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$,${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$,${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$, etc. In this question, we will make use of the identity ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)$ to factorise the given expression.
Step 1: Express in the form of ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz$.
We have $27={{3}^{3}}$
Hence $27{{x}^{3}}={{3}^{3}}{{x}^{3}}$
We know that ${{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}$
Hence, we have
$27{{x}^{3}}={{\left( 3x \right)}^{3}}$
Also, 9xyz = 3(3x)(y)(z).
Hence the expression becomes
${{\left( 3x \right)}^{3}}+{{y}^{3}}+{{z}^{3}}-3\cdot 3x\cdot y\cdot z$
Using ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz \right)$, we get
$27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz=\left( 3x+y+z \right)\left( {{\left( 3x \right)}^{2}}+{{y}^{2}}+{{z}^{2}}-3xy-yz-3xz \right)$
We know that ${{a}^{n}}{{b}^{n}}={{\left( ab \right)}^{n}}$
Using, we get
$27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz=\left( 3x+y+z \right)\left( 9{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3xy-yz-3xz \right)$
which is the required factored form of the given expression.

Note:  It is important to note that the term ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx$ is non-negative. This can be observed by noting that ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-xz=\dfrac{1}{2}\left( {{\left( x-y \right)}^{2}}+{{\left( y-z \right)}^{2}}+{{\left( x-z \right)}^{2}} \right)$
Hence the term is non-negative since square of a number is non-negative
 Using the above property, the above expression can be further simplified as
$27{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-9xyz=\dfrac{1}{2}\left( 3x+y+z \right)\left( {{\left( 3x-y \right)}^{2}}+{{\left( y-z \right)}^{2}}+{{\left( 3x-z \right)}^{2}} \right)$