Question

Factorize ${{\left( p-q \right)}^{3}}+{{\left( q-r \right)}^{3}}+{{\left( r-p \right)}^{3}}$

Answer 1

Hint: We will use the formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)+3abc\] to simplify the terms given in the expression. Then we will simplify the obtained equation and take some common terms out to get the desired answer.

Complete step by step solution:
We have been given an expression ${{\left( p-q \right)}^{3}}+{{\left( q-r \right)}^{3}}+{{\left( r-p \right)}^{3}}$.
We have to factorize the given expression.
In the given expression we have to find the factors. For this let us first simplify the given expression using the algebraic identity.
Now, we know that we have an algebraic formula \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)+3abc\].
Let us assume that $\left( p-q \right)=a,\left( q-r \right)=b,\text{ and }\left( r-p \right)=c$
Now, applying the above formula to each term of the given expression we will get
\[\begin{align}
  & = {{\left( p-q \right)}^{3}}+{{\left( q-r \right)}^{3}}+{{\left( r-p \right)}^{3}} \\
 & = \left( p-q+q-r+r-p \right)\left( {{\left( p-q \right)}^{2}}+{{\left( q-r \right)}^{2}}+{{\left( r-p \right)}^{2}}-3pq\left( p-q \right) \right)+\left( {{q}^{3}}-{{r}^{3}}-\left( p-q \right)\left( q-r \right)-\left( q-r \right)\left( r-p \right)-\left( r-p \right)\left( p-q \right) \right)+3\left( p-q \right)\left( q-r \right)\left( r-p \right) \\
\end{align}\]
Now, simplifying the above obtained equation we will get
\[= 0\times \left( {{\left( p-q \right)}^{2}}+{{\left( q-r \right)}^{2}}+{{\left( r-p \right)}^{2}}-3pq\left( p-q \right) \right)+\left( {{q}^{3}}-{{r}^{3}}-\left( p-q \right)\left( q-r \right)-\left( q-r \right)\left( r-p \right)-\left( r-p \right)\left( p-q \right) \right)+3\left( p-q \right)\left( q-r \right)\left( r-p \right)\]
Now, when we multiply a term with zero we will get the resultant as zero, so simplifying the above obtained equation we will get
\[= 3\left( p-q \right)\left( q-r \right)\left( r-p \right)\]
Hence above is the required factorized form of the given expression.

Note: Alternatively we can directly apply the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$ if $x+y+z=0$ by assuming $\left( p-q \right)=x,\left( q-r \right)=y,\text{ and }\left( r-p \right)=z$. Then simplifying the obtained equation we will get the desired answer.
First we will check if $x+y+z=0$
Then by substituting the values we will get
$= p-q+q-r+r-p$
Now, simplifying the above obtained equation we will get
$= 0$
So we can directly apply the formula to the given expression then we will get
$\begin{align}
  & = {{\left( p-q \right)}^{3}}+{{\left( q-r \right)}^{3}}+{{\left( r-p \right)}^{3}} \\
 & = 3\left( p-q \right)+\left( q-r \right)+\left( r-p \right) \\
\end{align}$
Hence above is the required factorize form of the given expression.