Find the LCM of $6 (x^{2} + 2 x y - 3 y^{2}), 4 (x^{2} - 3 x y + 2 y^{2}), 8 (x^{2} + x y - 6 y^{2})$

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Solution

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

Factorise $6 (x^{2} + 2 x y - 3 y^{2})$ as follows:

$6 (x^{2} + 2 x y - 3 y^{2}) = 6 (x^{2} + 3 x y - x y - 3 y^{2}) = 6 [x (x + 3 y) - y (x + 3 y)] = (2 \times 3) (x - y) (x + 3 y)$

Now, factorise $4 (x^{2} - 3 x y + 2 y^{2})$ as follows:

$4 (x^{2} - 3 x y + 2 y^{2}) = 4 (x^{2} - 2 x y - x y + 2 y^{2}) = 4 [x (x - 2 y) - y (x - 2 y)] = (2 \times 2) (x - y) (x - 2 y)$

Finally, factorise $8 (x^{2} + x y - 6 y^{2})$ as follows:

$8 (x^{2} + x y - 6 y^{2}) = 8 (x^{2} + 3 x y - 2 x y - 6 y^{2}) = 8 [x (x + 3 y) - 2 y (x + 3 y)] = (2 \times 2 \times 2) (x - 2 y) (x + 3 y)$

Therefore, the least common multiple between the polynomials $6 (x^{2} + 2 x y - 3 y^{2})$ , $4 (x^{2} - 3 x y + 2 y^{2})$ and $8 (x^{2} + x y - 6 y^{2})$ is:

$\Rightarrow L C M = 2 \times 2 \times 2 \times 3 (x - y) \times (x + 3 y) \times (x - 2 y) = 24 (x - y) (x + 3 y) (x - 2 y)$

Hence, the $L C M$ is $24 (x - y) (x + 3 y) (x - 2 y)$ .

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