we can do by expanding the same but can use the simple method as
below
1 * 6 = 2 * 3
so $(x+1)(x+2)(x+3)(x+6) - 3x^2$
=
$(x+1)(x+6)(x+2)(x+6)- 3x^2$
= $(x^2 + 7x + 6)(x^2 + 5x + 6) -
3x^2$
= $((x^2 + 6x+ 6) + x)((x^2 + 6x + 6) -x) - 3x^2$
= $(x^2+ 6x
+ 6)^2 - x^2 - 3x^2$
= $(x^2 + 6x+6)^2 - 4x^2$
= $(x^2 + 6x + 6)-
(2x)^2$
= $(x^2 + 8x + 6) (x^2 + 4x+ 6)$
above 2 cannot be factored further
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