Suppose $a < 0 < b$. Then what is the result of:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = ?$
I have a solution but I can't be sure if I did a mistake, because I usually do! My solution:
Call $a = -c$ for some $0 < c $. Then,
$=\sqrt{(-c-b)^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(- (c+b))^2} + \sqrt[6]{ b^6 }$
$=\sqrt{(c+b)^2} + \sqrt[6]{ b^6 }$
$=(c+b) + b$
$= -a +2b$
That seems good to me. You should be careful that $\sqrt{(c+b)^2} = |c + b|$. But as $c$ and $b$ are both positive, $|c + b| = c+ b$.
You can actually be more direct:
$\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = |a-b| + |b|$.
As $b > 0$ we know $|b| = b$. As $a < 0$ and $b > 0$ then then $a- b < a < 0$. So $|a-b| = b -a$
So $\sqrt{(a-b)^2} + \sqrt[6]{ b^6 } = |a-b| + |b| = b -a + b = -a + 2b$.
But your reasoning was just fine. But it may be worth stating, you replaced $a$ with $-c$ to make all variables positive.
.....
And just to be thorough:
$|a - b| + |b| = $
1)$ a - b + b=a$; if $b \ge 0$ and $a-b \ge 0$ i.e. if $b \ge 0$ and $a \ge b$, i.e if $0 \le b \le a$.
2) $a - b - b = a - 2b$; if $b \le 0$ and $a-b \ge 0$ i.e. if $b \le 0$ and $a \ge b$. (whether $a$ is bigger or small or equal to zero doesn't matter so long as $a \ge b$.
3)$b-a + b = -a + 2b$ if $b\ge0$ and $a-b \le 0$ i.e. if $a \le b$ and $b \ge 0$ (whether $a$ is bigger or smaller or equal to zero doesn't matter; Note; your problem was a subset of this case.)
4) $b-a - b = -a$ if $b\le 0$ and $a-b \le 0$ i.e. if $a \le b \le 0$.
(Note: Those four cases are not mutually exclusive. If $b=0$ then 1) $|a-b| + |b| = a$ and 2) $|a-b| + |b| = a-2b = a$ are compatible, as are 3) $|a-b| + |b| = -a + 2b = -a$ and 4) $|a -b| + |b| = -a$.)
(If $a-b = 0$ i.e. $a= b$ then 1) $|a-b| + |b| = a$ and 3) $|a-b| +|b| = -a + 2b = -a + 2a = a$ are compatible, as are 2) $|a-b| + |b|=a-2b = a-2a = -a$ and 4) |a-b|+|b| = -a$.
(if $a-b =0$ and $b=0$ then all four are compatible $|a-b| + |b| = 0 =a=-a = a-2b = -a + 2b$.)
i would write $$\sqrt{(a-b)^2}+\sqrt[6]{b^6}=\sqrt{(b-a)^2}+\sqrt[6]{b^6}=b-a+b$$ this is the same as you
