Question

Find the LCM of the following: $3(a-1),2(a-1{)}^2,(a^2-1)$

• A. $6(a-1{)}^2(a+1)$
• B. $6(a+1{)}^2(a+1)$
• C. $6(a-1{)}^2(a-1)$
• D. None of these

Answer 1

Answer: (A)

$6(a-1{)}^2(a+1)$

We know that the least common multiple (LCM) is the smallest number or expression that is a common multiple of two or more numbers or algebraic terms.

We first factorize the given binomials $3(a-1),2(a-1{)}^2$ and $a^2-1$ as shown below:

$3(a-1)=3\times (a-1)2(a-1{)}^2=2\times (a-1)\times (a-1)a^2-1=a^2-1^2=(a-1)\times (a+1\left)\right(\because x^2-y^2=(x-y)(x+y))$

Now multiply all the factors, using each common factor only once, therefore, the LCM is:

$3\times 2\times (a-1)\times (a-1)\times (a+1)=6(a-1{)}^2(a+1)$

Hence, the LCM of $3(a-1),2(a-1{)}^2$ and $a^2-1$ is $6(a-1{)}^2(a+1)$.