Finding a real root in a cubic equation?

Question

I am trying to find a real root of the following cubic.

$x^3-6x^2+14x-15=0$

I did $x=y-\frac{b}{3a}$

which is $x=y+2$

I plugged my substitution to get the depressing cubic.

$(y+2)^3-6(y+2)^2+14(y+2)-15=0$

which is

$y^3+2y=3$

then I set up the method for solving.

$y=s-t$ , first

$(3st=2)$, second

$s^3-t^3=3$

So then for the second equation I got $s=\frac{2}{3t}$

Plugged it into my third and simplified

$t^6+3t^3-\frac{8}{27}$

then Use the qudratic set $u=t^3$

so then I get

$u^2+3u-\frac{8}{27}$

Using the quadratic I got $-\frac{3}{2}+\frac{\sqrt{59/27}}{2}$

as my u which is t^3

take the cube root and I got $t=\sqrt[3]{\frac{-3}{2}+\frac{\sqrt{59/27}}{2}}$

so the s is $s=\sqrt[3]{\frac{3}{2}+\frac{\sqrt{59/27}}{2}}$

and since y=s-t you can plug than back into my $x=y+2$ equation to find x.

But I am not sure if I did this right

Answer 1

Note that $x=3$ is a solution. We used the Rational Roots Theorem, since cubics in exercises often have a rational root. Exercises and real life tend to differ.

When the depressed cubic was reached (correctly) the root $y=1$ was available by inspection.