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Without finding the cubes, factorize:
${\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3}$

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Last updated date: 19th Jul 2024
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Answer
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Hint: Given polynomial is of degree $3$. Polynomials of degree $3$ are known as cubic polynomials. The given cubic polynomial consists of three whole cube expressions or terms. We must solve this without cubing the expressions individually.

Complete step-by-step solution:
For factorising the given cubic polynomial ${\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3}$ without calculating the cubes if the expression, we must know the algebraic identities involving the cubic terms and expressions.
We know an algebraic identity of the form of sum of cubes of three terms ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$.
So, we can use this algebraic identity in the expression ${\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3}$, where a is $\left( {x - 2y} \right)$, b is $\left( {2y - 3z} \right)$ and c is $\left( {3z - x} \right)$.
So, before substituting the value of a, b and c in the expression ${a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) + 3abc$, we can evaluate the bracket $\left( {a + b + c} \right)$ separately in order to simplify the expression.
So, $\left( {a + b + c} \right) = \left[ {\left( {x - 2y} \right) + \left( {2y - 3z} \right) + \left( {3z - x} \right)} \right]$
$ \Rightarrow \left( {a + b + c} \right) = \left[ {x - 2y + 2y - 3z + 3z - x} \right]$
Cancelling all the like terms with opposite signs, we get,
$ \Rightarrow \left( {a + b + c} \right) = 0$
Now, the value of the bracket $\left( {a + b + c} \right)$ is zero. Hence, the value of the entire right side of the equation ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$ is zero since the multiplication of any number with zero is always equals to zero.
So, we get, ${a^3} + {b^3} + {c^3} - 3abc = 0$ if $\left( {a + b + c} \right) = 0$.
Now, substituting the values of a, b and c, we get,
$ \Rightarrow {\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3} - 3\left( {x - 2y} \right)\left( {2y - 3z} \right)\left( {3z - x} \right) = 0$
Now, keeping all the whole cube terms in left side of the equation and shifting the rest of the terms in right side of the equation, we get,
$ \Rightarrow {\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3} = 3\left( {x - 2y} \right)\left( {2y - 3z} \right)\left( {3z - x} \right)$
Hence, the factored form of the cubic polynomial ${\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3}$ is $3\left( {x - 2y} \right)\left( {2y - 3z} \right)\left( {3z - x} \right)$.

Note: One must know the required algebraic identity so as to solve this problem. We must also learn the special case of the algebraic identity ${a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) + 3abc$ when the value of $\left( {a + b + c} \right)$ is zero. We should always check for the value of the expression $\left( {a + b + c} \right)$ before applying the algebraic identity.