Question

factorise: 64x cube + y cube -z cube + 12xyz

Answer 1

$$\begin{gathered} 64x^3+y^3-z^3+12xyz \\ ={\left(4x\right)}^3+{\left(y\right)}^3+{\left(-z\right)}^3-3\times 4x\times y\times \left(-z\right) \\ =\left(4x+y-z\right)\left[{\left(4x\right)}^2+{\left(y\right)}^2+{\left(-z\right)}^2-4x\times y-y\left(-z\right)-\left(-z\right)4x\right] \\ \left[a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right] \\ =\left(4x+y-z\right)\left(16x^2+y^2+z^2-4xy+yz+4zx\right) \end{gathered}$$