How do i solve the cubic equation?

Question

$$x^3 - 3x^2 - 3x +2 = 0$$

The rational root test does not work; there are no rational roots.

Answer 1

Here is a step by step procedure.

First shift $x$ so there is no quadratic term. To do this we shift by 1/3 the quadratic term (assuming leading coefficient is 1) This give $$ x=y+1 \Rightarrow y^3-6\,y-3 =0 $$ We now reduce the cubic by the substitution $y = A(z+1/z)$ $$ y = A(z+1/z) \Rightarrow \frac{z^6\,A^3+3\,z^4\,A^3+3\,z^2\,A^3+A^3-6\,z^4\,A-6\,z^2\,A-3\,z^3}{z^3}=0 $$ We now pick $A \neq 0$ so that $Z^4$ term is zero. This requires $$ A(A^2-2)=0$$ So pick $A=\sqrt{2}$ and the equation becomes $$2^{{{3}\over{2}}}\,z^6-3\,z^3+2^{{{3}\over{2}}}=0$$ This is a quadratic in $z^3$. So set $z=u^{1/3}$ to get $$ z=u^{1/3} \Rightarrow 2^{{{3}\over{2}}}\,u^2-3\,u+2^{{{3}\over{2}}}=0$$

Solve this quadratic for $u$. You will have two roots. Use them to find $z$. Recall each number will have 3 cube roots, but you will have only 3 possible values and not 6.

Once you have $z$, calculate $y$ and then finally $x$.

Answer 2

There plainly there are no rational roots but the function does have inflexion points at $x=3(1\frac{+}{-} \sqrt2)$ subbing this into the original equation yields a positive and negative y value , respectively. So you may be able to find your root by graphing.