Question

Find the LCM of $4x^3+4x^2-x-1,8x^3-1,8x^2-2x-1$

Answer 1

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

Factorise $4x^3+4x^2-x-1$ as follows:

$4x^3+4x^2-x-1=4x^2(x+1)-1(x+1)=(4x^2-1)(x+1)=\left[\right(2x{)}^2-1^2\left]\right(x+1)=(2x+1\left)\right(2x-1\left)\right(x+1)$ (using identity $a^2-b^2=(a+b)(a-b)$)

Now, factorise $8x^3-1$ as follows:

$8x^3-1=(2x{)}^3-1^3=(2x-1\left)\right[(2x{)}^2+1+2x]=(2x-1)(4x^2+1+2x)$

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

Finally, factorise $8x^2-2x-1$ as follows:

$8x^2-2x-1=8x^2-4x+2x-1=4x(2x-1)+1(2x-1)=(4x+1)(2x-1)$

Therefore, the least common multiple between the polynomials $4x^3+4x^2-x-1$, $8x^3-1$and $8x^2-2x-1$ is:

$LCM=(2x+1)\times (2x-1)\times (x+1)\times (4x^2+1+2x)\times (4x+1)$
$=(x+1)(2x+1)(4x+1)\left[\right(2x-1\left)\right(4x^2+1+2x\left)\right]$

$=(x+1)(2x+1)(4x+1)(8x^3-1)$ (using identity $a^3-b^3=(a-b)(a^2+b^2+ab)$)

Hence, the $LCM$ is$(x+1)(2x+1)(4x+1)(8x^3-1)$.