6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$.

6.177. $\sqrt[3]{x+7}-\sqrt{x+3}=0$.
## Solution.

Domain of definition: $x+3 \geq 0, x \geq-3$.

From the condition $\sqrt[3]{x+7}=\sqrt{x+3}$ and raising both sides to the sixth power, we get $(x+7)^{2}=(x+3)^{3} \Leftrightarrow x^{2}+14 x+49=x^{3}+9 x^{2}+27 x+27 \Leftrightarrow$

$\Leftrightarrow x^{3}+8 x^{2}+13 x-22=0$. By checking, we find that $x_{1}=1$, since $1+8+13-22=0$. We divide the left side of the equation by $x-1$:

$\frac{x^{3}+8 x^{2}+13 x-22}{x-1}=x^{2}+9 x+22$.

Then $(x-1)\left(x^{2}+9 x+22\right)=0$.

From this, $x-1=0, x_{1}=1$, or $x^{2}+9 x+22=0$, no solutions exist since $D<0$. Answer: $x=1$.