Question

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

Answer 1

Hint: In this type of question we can apply the identity of $ {\left( {a - b} \right)^3} $ on the terms of the given expression $ {\left( {q - r} \right)^3},{\left( {r - p} \right)^3},{\left( {p - q} \right)^3} $ . Then simplify the expression till it becomes in multiples of linear equations (equation having degree 1).

Complete step-by-step answer:
The given expression for factorisation is: $ {p^3}{\left( {q - r} \right)^3} + {q^3}{\left( {r - p} \right)^3} + {r^3}{\left( {p - q} \right)^3} $
In this expression we see the terms $ {\left( {q - r} \right)^3},{\left( {r - p} \right)^3},{\left( {p - q} \right)^3} $ , and $ {p^3},{q^3},{r^3} $ have the same exponent and they are in multiple to each other.
So we know that, when two different values in multiple to each other and having same exponent, then their exponent become common to both terms and the values get multiply,
 $ {a^m}{b^m} = {(ab)^m} $
Applying this identity over the given expression we get,
 $ \Rightarrow {\left( {p\left( {q - r} \right)} \right)^3} + {\left( {q\left( {r - p} \right)} \right)^3} + {\left( {r\left( {p - q} \right)} \right)^3} $
Now taking the each term of the expression as variable then we get the expression as:
 $
\Rightarrow a = \left( {p\left( {q - r} \right)} \right)...................(i)\\
\Rightarrow b = \left( {q\left( {r - p} \right)} \right)...................(ii)\\
\Rightarrow c = \left( {r\left( {p - q} \right)} \right)...................(iii)
 $
Adding the equation (i), (ii) and (iii) we get,
 $ \Rightarrow a + b + c = \left( {p\left( {q - r} \right)} \right) + \left( {q\left( {r - p} \right)} \right) + \left( {r\left( {p - q} \right)} \right) $
By solving the expression we get,
 $ \Rightarrow a + b + c = pq - pr + qr - qp + rp - rq $
We found that all the given terms are in positive as well in negative form so, all get cancel then we get,
 $ a + b + c = 0 $
And we know that,
 $ {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $ (Identity)
Here the value of the expression \[a + b + c\] is 0. So after substituting the value in the identity then we get,
 $ {a^3} + {b^3} + {c^3} - 3abc = 0 \times \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) $
The value 0 is in multiple with expression of left-hand side make the whole expression 0 so we get the identity as:
 $
\Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0\\
\Rightarrow {a^3} + {b^3} + {c^3} = 3abc
 $
Substituting the value of a, b and c in the identity then we get,
 $ \Rightarrow {\left( {p\left( {q - r} \right)} \right)^3} + {\left( {q\left( {r - p} \right)} \right)^3} + {\left( {r\left( {p - q} \right)} \right)^3} = 3p\left( {q - r} \right) \times q\left( {r - p} \right) \times r\left( {p - q} \right)\\
{p^3}{\left( {q - r} \right)^3} + {q^3}{\left( {r - p} \right)^3} + {r^3}{\left( {p - q} \right)^3} = 3pqr\left( {q - r} \right)\left( {r - p} \right)\left( {p - q} \right) $
This is the required factorise form the given expression. $ 3pqr\left( {q - r} \right)\left( {r - p} \right)\left( {p - q} \right) $ are the factors of the given expression.

Note: In this type of problem, we can also find the factorise of the expression by using the long division method still we will get the same result. Using algebraic identities we just compare the expression with identities and expand it.