Isolate the root expression:
![\sqrt[3]{x+1}+2=0\implies\sqrt[3]{x+1}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%2B2%3D0%5Cimplies%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2)
Take the third power of both sides:
![\sqrt[3]{x+1}=-2\implies(\sqrt[3]{x+1})^3=(-2)^3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B1%7D%3D-2%5Cimplies%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3)
Simplify:
![(\sqrt[3]{x+1})^3=(-2)^3\implies x+1=-8](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx%2B1%7D%29%5E3%3D%28-2%29%5E3%5Cimplies%20x%2B1%3D-8)
Isolate and solve for

:

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

, we have
![\sqrt[3]{-9+1}=\sqrt[3]{-8}=\sqrt[3]{(-2)^3}=-2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-9%2B1%7D%3D%5Csqrt%5B3%5D%7B-8%7D%3D%5Csqrt%5B3%5D%7B%28-2%29%5E3%7D%3D-2)
as required.
Answer:
Step-by-step explanation:
36 for 8 because 24 for 5$ isnt nearly as good as 36 for 8$
Answer:
4
Step-by-step explanation:
List all the factors of both numbers:
32: 1, 2, 4, 8, 16, 32
60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Now we look for the greatest number that is a factor of both 32 and 60.
Answer: 4
Answer:
x=2
y=-8
Step-by-step explanation:
6x+3y=-12
-6x-2y=4
3y=-12-6x
y=(-12-6x)/3
y=-4-2x
substitute y
-6x-2(-4-2x)=4
-6x+8+4x=4
-2x=-4
x=2
6(2)+3y=-12
12+3y=-12
3y=-24
y=-8
check:
6(2)+3(-8)=-12
12-24=-12
-6(2)-2(-8)=4
-12=16=4