Question

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

Answer 1

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

Factorise $21(x-1{)}^2$ as follows:

$21(x-1{)}^2=(3\times 7\left)\right(x-1\left)\right(x-1)$

Now, factorise $35(x^4-x^2)$ as follows:

$35(x^4-x^2)=35x^2(x^2-1)=(5\times 7)x^2(x-1)(x+1)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Finally, factorise $14(x^4-x)$ as follows:

$14(x^4-x)=14x(x^3-1)=14x\left[\right(x-1\left)\right(x^2+1^2+x)=(2\times 7\left)x\right(x-1\left)\right(x^2+x+1)$

(using identity $a^3-b^3=(a-b)(a^2+b^2+ab)$)

Therefore, the least common multiple between the polynomials $21(x-1{)}^2$, $35(x^4-x^2)$and $14(x^4-x)$ is:

$LCM=2\times 3\times 5\times 7\times x^2\times (x+1)\times (x-1)\times (x-1)\times (x^2+x+1)=210x^2\left[\right(x+1\left)\right(x-1\left)\right]\left[\right(x-1\left)\right(x^2+x+1\left)\right]$

$=210x^2(x^2-1)(x^3-1)$

(using identities $a^3-b^3=(a-b)(a^2+b^2+ab)$ and $a^2-b^2=(a+b)(a-b)$)

Hence, the $LCM$ is $210x^2(x^2-1)(x^3-1)$.