Find the least common multiple of 9x^2-16 and 3x^2+x-4
\(9x^2-16=(3x-4)(3x+4)\)
\(3x^2+x-4\\=3x^2+4x-3x-4\\=x(3x+4)-1(3x+4)\\=(x-1)(3x+4)\)
So the lowest common multiple is
\((3x-4)(3x+4)(x-1)\\ =(9x^2-16)(x-1)\\ =9x^3-16x-9x^2+16\\ =9x^3-9x^2-16x+16\\ \)
