What is the answer to cube root of x+1}=\sqrt[3]{x^2-5 ?

The answer to cube root of x+1}=\sqrt[3]{x^2-5 is x=3,x=-2

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cube root of x+1}=\sqrt[3]{x^2-5

x = 3, x = - 2

Solution steps

Take both sides of the equation to the power of

3 : x + 1 = x^{2} - 5

x + 1 = x^{2} - 5

Solve

x + 1 = x^{2} - 5 : x = 3, x = - 2

x = 3, x = - 2

Verify Solutions:

x = 3

True

, x = - 2

True

The solutions are

x = 3, x = - 2

4 x + 25 - 5 = - 2

6 x + 46 - 18 = - 16

x, t x = 2 - x 2 - 9

2 x + 21 - 17 = - 14

2 x + 7 + 3 x + 10 = 2

What is the answer to cube root of x+1}=\sqrt[3]{x^2-5 ?
The answer to cube root of x+1}=\sqrt[3]{x^2-5 is x=3,x=-2
Solve for x: cube root of x+1}=\sqrt[3]{x^2-5
The solution is x=3,x=-2