Question

In two or more complete sentences, explain how to solve the cube root equation, %5D%7Bx%2B1%7D%20%2B2%3D0" id="TexFormula1" title=" \sqrt[3]{x+1} +2=0" alt=" \sqrt[3]{x+1} +2=0" align="absmiddle" class="latex-formula">

Answer 1

Isolate the root expression:

$\sqrt[3]{x+1}+2=0\implies\sqrt[3]{x+1}=-2$

Take the third power of both sides:

$\sqrt[3]{x+1}=-2\implies(\sqrt[3]{x+1})^3=(-2)^3$

Simplify:

$(\sqrt[3]{x+1})^3=(-2)^3\implies x+1=-8$

Isolate and solve for $x$:

$x=-9$

Since the cube root function is bijective, we know this won't be an extraneous solution, but it doesn't hurt to verify that this is correct. When $x=-9$, we have

$\sqrt[3]{-9+1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2$

as required.