Question

If x = 2y + 6, find the value of (x³ − 8y³ − 36xy − 216)

Answer 1

$$\begin{gathered} x^3-8y^3-216-36xy \\ ={\left(x\right)}^3+{\left(-2y\right)}^3+{\left(-6\right)}^3-3\times x\times \left(-2y\right)\times \left(-6\right) \\ =\left[x+\left(-2y\right)+\left(-6\right)\right]\left(x^2+{\left(-2y\right)}^2+{\left(-6\right)}^2-x\left(-2y\right)-\left(-2y\right)\times \left(-6\right)-\left(-6\right)x\right) \\ =\left(x-2y-6\right)\left(x^2+4y^2+36+2xy-12y+6x\right) \\ =\left[2y+6-2y-6\right]\left(x^2+4y^2+36+2xy-12y+6x\right)\left(\mathrm{Given}:x=2y+6\right) \\ =0\times \left(x^2+4y^2+36+2xy-12y+6x\right) \\ =0 \end{gathered}$$