# In two or more complete sentences, explain how to solve the cube root equation, yr -1 + 2 = 0.

###### Question:

## Answers

[tex]\sqrt[3]{x} - 1 + 3 = 0 \\ \sqrt[3]{x} + 2 = 0 \\ \sqrt[3]{x} = - 2 \\ x = { - 2}^{3} \\ x = - 8[/tex]

That's the result of this equation. I hope it will help you. Just do it like basic equation until you come to (3rd root of x is equal to -2). Then get rid of root. Just put the exponent to right side and you'll get x=-2^3 and just calculate it. Final result is -8

[tex]x=-9[/tex]

Step-by-step explanation:

Here we have , We need to solve equation ^3√x+1 +2=0 or , [tex]\sqrt[3]{x+1}+2=0[/tex] .Let's solve this :

⇒ [tex]\sqrt[3]{x+1}+2=0[/tex]

⇒ [tex](\sqrt[3]{x+1}+2)-2=0-2[/tex]

⇒ [tex]\sqrt[3]{x+1}=-2[/tex]

Cubing both sides

⇒ [tex](\sqrt[3]{x+1})^3=(-2)^3[/tex]

⇒ [tex](\sqrt[3]{x+1})^3=-8[/tex]

⇒ [tex](x+1)^{(\frac{1}{3})^3}=-8[/tex]

⇒ [tex]x+1=-8[/tex]

⇒ [tex]x=-9[/tex]

Therefore , [tex]x=-9[/tex] .

PEMDAS

Step-by-step explanation: You wanna add 1 to both sides and get 3√x = 1. Then you wanna multiply 3 by 3√x which leaves you with x cause it canceled out. But remember to multiply 3 on both sides so x = 1 sq rt 3. Which is just 1.

x = 1

∛x-1+3 = 0

Add -3 to both sides to give ∛x-1 = -3.

Now cube both sides to give x - 1 = -27. Add 1 to both sides to find x, which is -26.

(b) Along the straight line from (0, 0, 0) to (0, 1, 1), x = 0, y = 0, dx = 0, dy = 0, while z ... 6(1)(1)}dx + {2(1)+3x(1)} + {1 – 4x(1)(19°30 = S (3x' – 6)dx=-5 ... (c) The straight line joining (0,0,0) and (1, 1, 1) is given in parametric form by x = 1, y = 1, z = 1. ... From this it follows that the line integral around any closed path in R is zero, ...

Step-by-step explanation:

[tex]x= -7[/tex]

Step-by-step explanation:

We need to solve the following root equation:

[tex]\sqrt[3]{x-1} + 2 = 0[/tex]

1. First, we need to add minus two to both sides of the equation:

[tex]\sqrt[3]{x-1}= -2[/tex]

2. Elevate each side of the equation to the power of 3:

[tex]x-1= -8[/tex]

3. Solve for 'x':

[tex]x= -8 + 1[/tex]

[tex]x= -7[/tex]

Therefore, the result is:

[tex]x= -7[/tex]✔️

Explanation:

The first step is to figure out what the equation is. When unconventional math symbols are used, and when there are no grouping symbols identifying operands, that can be the most difficult step. Here, we think the equation is supposed to be ...

[tex]\sqrt[3]{x+1}-2=0[/tex]

It usually works well in radical equations to isolate the radical. Here that would mean adding 2 to both sides of the equation, to undo the subtraction of 2.

[tex]\sqrt[3]{x+1}=2[/tex]

Now, it is convenient to raise both sides of the equation to the 3rd power.

[tex]x+1=2^3=8[/tex]

Finally, we can isolate the variable by undoing the addition of 1. We accomplish that by adding -1 to both sides of the equation.

[tex]x=7[/tex]

The equation is solved. The solution is x = 7.

Hello from MrBillDoesMath!

x = -1/27

Discussion:

3 x^(1/3) -1 +2 = 0 =>

3 x^(1/3) +1 = 0 => (as -1 + 2 = 1).

3 x^(1/3) + 1 - 1 = 0 - 1 => Add -1 to both sides

3 x^(1/3) = -1 => (divide both sides by 3)

(3/3) x^(1/3) = -1/3 =>

x^(1/3) = -1/3 =>

x = (-1/3) ^3 =>

x = -1/27

Thank you,

MrB

hi

Step-by-step explanation:

we have

[tex]\sqrt[3]{x} -1+2=0[/tex]

Step 1

Solve like terms

[tex]\sqrt[3]{x}+(-1+2)=0[/tex]

[tex]\sqrt[3]{x}+(1)=0[/tex]

Step 2

Subtract 1 from both sides of the equation

[tex]\sqrt[3]{x}+1-1=0-1[/tex]

[tex]\sqrt[3]{x}=-1[/tex]

Step 3

Cube both sides of the equation in order to remove the radical on the left side

[tex](\sqrt[3]{x})^{3}=(-1 )^{3} \\ x=-1[/tex]

therefore

the answer is

[tex]x=-1[/tex]