Question

The identity \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\] can be easily verified by expanding the right hand side.
Using the above identity (or otherwise) answer the following:
Factorize the expression:
\[{\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3}\]
(a) \[\left( {p + q + r} \right)\left( {{p^2} + {q^2} + {r^2} - 2\left( {pq + qr + pq} \right)} \right)\]
(b) \[\left( {p + q + r} \right)\left( {2{p^2} + 2{q^2} + 2{r^2} - 3\left( {pq + qr + rp} \right)} \right)\]
(c) \[3\left( {p + 2q - 3r} \right)\left( {q + 2r - 3p} \right)\left( {r + 2p - 3q} \right)\]
(d) None of these

Answer 1

Hint: Here, we need to compute the given expression. We will rewrite the given algebraic identity taking the sum of the numbers as zero. Then, we will simplify the expression using the identity to find the required value of the given expression.

Formula Used:
\[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\]

Complete step-by-step answer:
The given algebraic identity is \[{a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)\].
Adding \[3abc\] to both sides of the algebraic identity, we get
\[\Rightarrow {a^3} + {b^3} + {c^3} - 3abc + 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) + 3abc \\
\Rightarrow {a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) + 3abc \\ \]
Suppose that the sum of the three numbers is 0.
Therefore, we have
\[a + b + c = 0\]
Substituting \[a + b + c = 0\] in the identity, we get
\[ \Rightarrow {a^3} + {b^3} + {c^3} = 0\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) + 3abc\]
We know that when any number or expression is multiplied by 0, the result is 0.
Therefore, we get
\[ \Rightarrow {a^3} + {b^3} + {c^3} = 0 + 3abc\]
Adding the terms, we get
\[ \Rightarrow {a^3} + {b^3} + {c^3} = 3abc\]
Now, we will find the value of the given expression.
The given expression is \[{\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3}\].
Let \[a = p + 2q - 3r\], \[b = q + 2r - 3p\], and \[c = r + 2p - 3q\].
We will check the sum of these numbers.
Substituting \[a = p + 2q - 3r\], \[b = q + 2r - 3p\], and \[c = r + 2p - 3q\] in the expression \[a + b + c\], we get
\[ \Rightarrow a + b + c = p + 2q - 3r + q + 2r - 3p + r + 2p - 3q\]
Adding the like terms in the expression, we get
\[ \Rightarrow a + b + c = 3p + 3q - 3r + 3r - 3p - 3q\]
Pairing the terms of the expression, we get
\[ \Rightarrow a + b + c = \left( {3p - 3p} \right) + \left( {3q - 3q} \right) + \left( {3r - 3r} \right)\]
Subtracting the terms in the parentheses, we get
\[ \Rightarrow a + b + c = 0 + 0 + 0\]
Therefore, we get
\[ \Rightarrow a + b + c = 0\]
We have proved using the algebraic identity that if \[a + b + c = 0\], then \[{a^3} + {b^3} + {c^3} = 3abc\].
Since \[a + b + c = \left( {p + 2q - 3r} \right) + \left( {q + 2r - 3p} \right) + \left( {r + 2p - 3q} \right) = 0\], we can use \[{a^3} + {b^3} + {c^3} = 3abc\].
Substituting \[a = p + 2q - 3r\], \[b = q + 2r - 3p\], and \[c = r + 2p - 3q\] in the equation \[{a^3} + {b^3} + {c^3} = 3abc\], we get
\[ \Rightarrow {\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3} = 3\left( {p + 2q - 3r} \right)\left( {q + 2r - 3p} \right)\left( {r + 2p - 3q} \right)\]
Therefore, we get the value of the expression \[{\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3}\] as \[3\left( {p + 2q - 3r} \right)\left( {q + 2r - 3p} \right)\left( {r + 2p - 3q} \right)\].
Thus, the correct option is option (c).

Note: We supposed that the sum of the three numbers in the identity is 0. This is because if the sum of the three numbers is 0, then the right hand side is equal to 0. The sum of the three numbers given in the expression \[{\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3}\] is 0. By rewriting the identity when the sum of the three numbers \[a + b + c = 0\], we have formed a simplified identity to find the value of \[{\left( {p + 2q - 3r} \right)^3} + {\left( {q + 2r - 3p} \right)^3} + {\left( {r + 2p - 3q} \right)^3}\].