Solve $(x-1)^3 = 1$ I'm in 13th class (I think in the UK it is called "Upper Sixth Form") and in our math-book there is the following question:
At which point does the graph $$g(x) = \frac{1}{(x-1)^{2}}$$ have a gradient of $-2$?
So I built the derivation, which is $$g'(x) = \frac{-2}{(x-1)^{3}}$$
For g' to be $-2$, the following term has to be true:
$$(x-1)^{3} = 1$$
This is my problem, I have no idea how to solve this equation. We have not yet had any formula for solving terms with 3 as exponent. But since it is in our book there has to be some easier way.
Thanks!
 A: $(x-1)^3=1$ is the same as $x-1=1^{\frac{1}{3}}$
A: You're solving $(x-1)^3-1=0$. Remember the formula:
$$a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$$ $$(x-1)^3-1=((x-1)-1)\left((x-1)^2+(x-1)+1\right)=0$$
This holds if and only if either $(x-1)-1=0$ or $(x-1)^2+(x-1)+1=0$, both you know how to solve.

Or apply the bijective function $f(x)=\sqrt[3]{x}$ on both sides: $$(x-1)^3=1\iff \sqrt[3]{(x-1)^3}=\sqrt[3]{1}\iff x-1=1\iff x=2$$
Or notice the function $f(x)=(x-1)^3$ is strictly increasing, and $x=2$ is a solution, so $x=2$ is the only solution.
A: The way to solve cubics with arbitrary coefficients might be difficult (tedious) but your book assumes solving some easily factorable and simple termed cubics has been covered and I'm sure it has. 
Like solving any equation take inverses of both sides of the equation to isolate the x.
$(x - 1)^3 = 1$
$(x - 1) = \sqrt[3]1 = 1$
$x = 2$.  (That's one root.  The other two are complex.)
The difficulty is only when finding the proper procedure to isolate the x isn't clear.  That isn't the case here. The book may not have covered how to isolate x in cubics in general (which isn't straightforward) but this isn't a a cubic in general.
