This question was uploaded on 13/01/22 on social media accounts.
Solution:
Given equation:
\[\color{blue} {x^3=32(x+2)=32x+64}\]
\[\Rightarrow \color{blue} {x^3-16x=16x+64}\]
\[\Rightarrow \color{blue} {x(x^2-16)=16(x+4)}\]
\[\Rightarrow \color{blue} {x(x-4)\color{Red}{(x+4)}=16\color{red}{(x+4)}}\]
\[\Rightarrow \color{blue} {x(x-4)=\color{#228b22}{x^2-4x=16}}\]
In this, (x+4) term is canceled,
\[\Rightarrow \color{#ff1694}{x+4=0}\Rightarrow \color{#ff1694}{x=-4}\]
\[\Rightarrow \color{#228b22}{x^2-4x=32}\]
\[\Rightarrow \color{#228b22}{x^2-4x=\left\{16,32\right\}}\]
