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Find the LCM of $21{(x - 1)^2},$ $35({x^4} - {x^2}),$ $14({x^4} - x)$

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Last updated date: 20th Jun 2024
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Answer
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Hint: The LCM (Least Common Multiple) of the given expressions is the smallest expression that is divisible by each of the given expressions. First express each of the given expressions $21{(x - 1)^2},$ $35({x^4} - {x^2}),$ $14({x^4} - x)$ as the product of its factors including the prime factors where so ever applicable. Find out the common factors from each of the given expressions. Take the product of each of the common factors with the highest powers by simplifying the expressions. The resultant is the LCM of the given expression.


Complete step by step solution
LCM of $21{(x - 1)^2},$ $35({x^4} - {x^2}),$ $14({x^4} - x)$ is equal to –
Taking factorization of all the three given expressions –
$21{(x - 1)^2} = 3 \times 7 \times (x - 1) \times (x - 1)$ ................................................................(A)
[Because, we know that - ${(a - b)^2} = (a - b)(a - b)$ ]
Similarly,
 $35({x^4} - {x^2}) = 5 \times 7 \times {x^2}({x^2} - 1)$
$35({x^4} - {x^2}) = 5 \times 7 \times {x^2}(x - 1)(x + 1)$ .............................................................(B)
 [Apply the formula for the difference of two squares- ${a^2} - {b^2} = (a - b)(a + b)$ ]
Similarly, taking the factorisation of –
$14({x^4} - x) = 2 \times 7 \times x({x^3} - 1)$
[Now, as per the property of the difference of cubes- ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$ ]
$14({x^4} - x) = 2 \times 7 \times x \times (x - 1)({x^2} + x + 1)$ .....................................................(C)
Now, by observing all the equations (A), (B) and (C)
Take common factors from all the three given expressions –
Therefore, LCM $ = 7(x - 1)$
Hence, LCM of $21{(x - 1)^2},$ $35({x^4} - {x^2}),$ $14({x^4} - x)$ $ = 7(x - 1)$and is the required answer.

Note: To solve the least common multiple (LCM) of any given expressions, the concept of prime numbers should be clear. Prime numbers are the natural numbers greater than $1$ and which are not the product of any two smaller natural numbers. $1$ is neither prime nor composite. For Example: $2,3,5,7$$,.....$ $2$ is the prime number as it can have only $2$ factors. The number $1$ (one) and the number itself that is $2$. Hence, the factors of $2 = 2 \times 1$