Solutions of $\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1$

Question

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

I get that $-3$ as a solution, but apparently 1 is as well a solution, and I don't see a mechanism through which I could find it. Any help would be appreciated.

Answer 1

we have $$\sqrt{x+4+2\sqrt{x+3}}-(x^2+4x+3)^{1/3}=1 $$ let us make a change of variable $u = x + 3 \ge 0, x = u - 3.$ with that we have $$\sqrt{u+1+2\sqrt u}-(u(u-2))^{1/3}=1 \to 1+\sqrt u=(u(u-2))^{1/3}+1$$ this gives us $$\sqrt u = (u(u-2))^{1/3}\tag 1 $$

now, exponentiating $(1)$ implies $$ 0=u^3 - u^2(u-2)^2 \to 0=u^2(u^2 -5u + 4)=u^2(u-4)(u-1) $$

subbing $u = 1,$ in $(1),$ we have $1 = -1$ therefore is an extraneous solution and needs to be rejected.

the roots are $u = 0, u = 4$ are solutions of $1.$ this translates to $$x = -3, x = 1. $$