I realise that the radicands are perfect cubes and rewrite
$$\sqrt{(x-3)^2}-\sqrt{(x+3)^2}\Leftrightarrow|x-3|-|x+3|.$$ The possible answers are
a) $6$
b) $0$
c) $-12x$
d) None of the above is correct for all $x<0$.
The term $|x-3|$ tells me, geometrically, that $x$ is $3$ units away from $3$ on the $x$-axis and $|x+3|=|x-(-3)|$ tells me that $x$ is $-3$ units from the $x$-axis. Does this mean that we have $|x-3|-|x+3|=3-(-3)=6?$
On
I realise that the discriminants are perfect cubes and rewrite
Did you perhaps mean the radicands are perfect squares?
Anyway, since $x < 0$, then $x - 3 < -3 < 0$ and so $|x-3| = -(x-3) = 3 - x$ for all $x < 0$.
Also, $x + 3 < 3$, which means $x+3$ can be positive or negative (or zero). So, $|x+3| = x+3$ if $-3 \le x < 0$, and $|x+3| = -(x+3) = -x-3$ if $x < -3$.
So we have two cases:
On
HINT: we have three cases: 1)$x\geq 3$ then we get $$x-3-x-3=-6$$ 2) $$-3\le x<3$$ then we have $$-x+3-x-3=-2x$$ 3) $$x<-3$$ then we get $$-x+3+x+3=6$$
On
If $x<0$, then $x-3<0$ and therefore $|x-3|=-(x-3)=3-x$. If it turns out that $x+3<0$, then $|x+3|=-x-3$ and then $|x-3|-|x+3|=6$. But if $x+3>0$, then $|x+3|=x+3$ and, in this case, $|x-3|-|x+3|=-2x$.
Can we have $x<0$ and $x+3<0$? Yes: take $x=-4$. And can we have $x<0$ and $x+3>0$? Yes: take $x=-2$. Therefore, the answer is D.
On
$\sqrt{(x-3)^2}-\sqrt{(x+3)^2}=|x-3|-|x+3|$
(NOT $\iff$... The phrases $\sqrt{(x-3)^2}-\sqrt{(x+3)^2}$ as $|x-3|-|x+3|$ are not a sentences or a statements. They are simply values. This doesn't make any more sense than $5 \iff 7-2$. What about 5? What about $7-2$?)
... and ....
$|x-3|-|x+3| = \pm (x-3) - \pm (x+3)$
Case 1: $|x-3| = x -3$ and $|x+3| = x+3$.
That means $x-3 \ge 0$ and $x \ge 3 > 0$. That's a contradiction this is not possible.
So $|x-3|-|x+3| \ne (x-3) - (x+3)$.
Case 2: $|x-3| = x-3$ and $|x +3| = -(x+3)$. That's a contradiction for the exact same reason. $|x-3| \ne x -3$.
Case 3: $|x-3| = -(x-3)$ and $|x+3| = x+3$.
Then $x-3 < 0$ and $x < 3$. That is true. And $x+3 \ge 0$ so $x \ge -3$. That is .... possible. But not an absolute certainty.
IF $-3 \le x < 0$ then $|x-3| -|x+3| = -(x-3) -(x+3) = -2x$.
But this might not be true if $x < -3$.
Case 4: $|x-3| = -(x-3)$ and $|x+3| = -(x+3)$
If so then $x-3 < 0$ and $x < 3$ which is true. $x+3 < 0$ and $x < -3$ which is possible.
IF $x < -3$ then $|x-3| - |x+3| = -(x-3) -(-(x+3)) = (3-x) + (x+3) = 6$
...
So we have two possibilities.
Either $0 > x \ge -3$ and $|x-3| - |x+3| = -2x$. OR. $x < -3$ and $|x-3| - |x+3| = 6$.
There is no single one answer for all $x < 0$.
So the answer is D.
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Note if we didn't specify that $x < 0$ we'd have the possibilties that
If $0 \le x < 3$
The $|x-3| = -(x-3) = 3-x$ and $|x+3| = x+3$ so $|x-3| - |x+3| = (3-x)-(x+3)=-2x$
If $x \ge 3$ then $|x-3| - |x+3| = (x-3) - (x+3) = -6$.
So if we didn't have $x < 0$ we'd have three case.
If $x < -3$ then $|x-3| - |x+3|=6$.
If $-3 \le x \le 3$ then $|x-3| - |x+3|=-2x$.
And if $x > 3$ then $|x-3| - |x+3| = -6$.
The three areas in question for this problem are $x\lt-3$, $-3\le x \le3$, and $x \gt 3$.
For $x\gt 3$ $$|x−3|−|x+3| = x-3 - (x+3) = -6$$ For $-3\le x \le3$ $$|x−3|−|x+3| =-(x-3)-(x+3) = -x+3-x-3 = -2x$$ For $x\lt-3$ $$|x−3|−|x+3| = -(x-3) - -(x+3) = -x + 3+x+3 = 6$$
Clearly, for $x \lt 0$, we have two equations, $y=-2x$ until $x=-3$ and $y = 6$ after $x=-3$. Therefore your answer is D.