Not sure how to find out what $x$ is...

Question

$$(x^2 -1)^3=27$$

So I’ve been thinking that $x$ could be found by:

$$(x^2 -1)^3=27$$

$$x^{2 \cdot 3} -1^3=27$$

$$x^6=27+1$$

$$x^6=28$$

$$x＝6^{\text{th}}\text{ root of }28$$

$$x=31.75$$

But apparently it appears to be incorrect, need help

Answer 1

Always unwrap. That's all there is to it (and being careful.

$(x^2 - 1)^3 = 27$ top layer of the wrapping is $something^3 = ourresult$. So we must unwrap $something^3$. The way to unwrap cube powers is but takeing cube roots.

So $\sqrt[3]{(x^2 - 1)^3} = \sqrt[3] {27} = 3$.

Now we have to be careful. Is the $\sqrt[3]{something^3}= something$ the only option. Is it possible for $something^3 = ourresult$ but NOT have $something = \sqrt[3]{ourresult}$? In this case, no $something^3 = ourresult$ if and only if $something = \sqrt[3]{ourresult}$.

So $x^2 - 1 = 3$.

Now we have $something -1 = ourresult$. So we must unwrap $-1$. The way to unwrap a subtract is to add.

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

Is adding a unique operation? Yes, is is.

So $x^2 = 4$

So now we have $something^2 = ourresult$. The top wrapping is a square. Then we to unwrap a square is to take a square root.

$\sqrt{x^2} = \sqrt{4} =2$.

But BE CAREFUL! This time squaring and square rooting is NOT unique and $x^2 = w$ does NOT mean $x = \sqrt{w}$. If $x^2 = w$ then maybe $x = \sqrt{w}$ but it is also possible that $x = -\sqrt{w}$.

So $\sqrt{x^2} = x$ or $-x$ so $\pm x = 2$ or $x = \pm 2$.

Answer 2

$$(a-b)^3\neq a^3-b^3$$ for all values $a$ and $b$. Take $a=2$ and $b=1$.

I think it's better to use the following way.

It's $$x^2-1=3$$ or $$x^2=4,$$ which gives the answer $$\{2,-2\}$$