Question

If $(x-1{)}^3=8$, what is the value of $(x+1{)}^2$ for real $x$?

Answer 1

Given,

$(x-1{)}^3=8$

or, $(x-1{)}^3=2^3$

The only real root is $x=1+2=3$. [ Other roots (complex roots) are $1+2\omega ,1+2{\omega }^2$ where $\omega$ is cube root of unity]

Now,

$(x+1{)}^2$

$=(3+1{)}^2$

$=16$.