Question

Find the LCM of $a^2-1,a^4-1,a^8-1$

Answer 1

We know that $LCM$ is the least common multiple.

Factorise $a^2-1$ as follows:

$a^2-1=(a-1)(a+1)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Now, factorise $a^4-1$ as follows:

$a^4-1=(a^2{)}^2-1^2=(a^2-1\left)\right(a^2+1)=(a-1\left)\right(a+1\left)\right(a^2+1)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Finally, factorise $a^8-1$ as follows:

$a^8-1=(a^4{)}^2-1^2=(a^4-1\left)\right(a^4+1)=[(a^2{)}^2-1^2](a^4+1)=(a^2-1)(a^2+1)(a^4+1)$

$=(a-1)(a+1)(a^2+1)(a^4+1)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Therefore, the least common multiple between the polynomials $a^2-1$, $a^4-1$and $a^8-1$ is:

$LCM=(a-1)(a+1)(a^2+1)(a^4+1)$

Hence, the $LCM$ is $(a-1)(a+1)(a^2+1)(a^4+1)$.