Question

Solve:
$x^3-4x^2-x+1=(x-2{)}^3$

Answer 1

$x^3-4x^2-x+1=(x-2{)}^3$

using identify $(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

we get

$x^3-4x^2-x+1=x^3-8-6x^2+12x$

$6x^2-4x^2-x-12x+1+8=0$

$2x^2-13x+9=0$

$x=\frac{+13\pm \sqrt{169-72}}{4}=\frac{13\pm \sqrt{97}}{4}$

Hence $x=\frac{13+\sqrt{97}}{4}$ & $\frac{13-\sqrt{97}}{4}$