Question

Find the cube of 3x-2y-4z

Answer 1

`(a-b)^3=a^3-3\cdot a^2\cdot b+3\cdot a\cdotb^2-b^3\\\\(a+b)^3=a^3+3\cdot a^2\cdot b+3\cdot a\cdotb^2+b^3\\\\--------------------\\\\(3x-2y-4z)^3=[3x-(2y+4z)]^3=\\\\(3x)^3-3\cdot(3x)^2\cdot(2y+4z)+3\cdot3x\cdot (2y+4z)^2-(2y+4z)^3=\\\\`


`=27x^3-3\cdot9x^2\cdot(2y+4z)+9x\cdot[(2y)^2+2\cdot2y\cdot4z+(4z)^2]-[(2y)^3+\\\\+3\cdot(2y)^2\cdot4z+3\cdot2y\cdot(4z)^2+(4z)^3]=\\\\=27x^3-54x^2y-108x^2z+9x(4y^2+16yz+16z^2)-(8y^3+12z\cdot4y^2+\\\\+6y\cdot16z^2+64z^3)=27x^3-54x^2y-108x^2z+36xy^2+144xyz+\\\\+144xz^2-8y^3-48y^2z-96yz^2-64z^3=\\\\=27x^3-8y^3-64z^3-54x^2y-108x^2z+36y^2x-48y^2z+\\\\+144z^2x-96z^2y+144xyz`

Answer 2

so the cube of (3x-2y-4z) is also
(3x-2y-4z)^3
(3x-2y-4z)(3x-2y-4z)(3x-2y-4z)

so we do the first two first
(3x-2y-4z)(3x-2y-4z)

(9x^2-12xy-24xz+4y^2+16yz+16z^2)
multiply this by (3x-2y-4z) and get
(27x^3-54yx^2-108zx^2+36xy^2+144xyz+144xz^2-8y^3-48zy^2-96yz^2-64z^3)