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

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Solution

We know that $L C M$ is the least common multiple.

Factorise $x^{3} + 8$ as follows:

$x^{3} + 8 = x^{3} + 2^{3} = (x + 2) (x^{2} + 2^{2} - 2 x)$ ............(using identity $a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)$ )
$= (x + 2) (x^{2} + 4 - 2 x)$

Now, factorise $x^{2} - 4$ as follows:

$x^{2} - 4 = x^{2} - 2^{2} = (x + 2) (x - 2)$ .........(using identity $a^{2} - b^{2} = (a + b) (a - b)$ )

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

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

Hence, the $L C M$ is $(x - 2) (x^{3} + 8)$ .

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