What is the answer to cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4 ?

The answer to cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4 is x=2,x=-2

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cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4

x = 2, x = - 2

Solution steps

Rewrite the equation with

and

x = u^{3} - 2

Solve

Substitute back

solve for

x

Solve

x = 2, x = - 2

Verify Solutions:

x = 2

True

, x = - 2

True

The solutions are

x = 2, x = - 2

3 x - 1 - 4 = 2

x - \frac{1}{2} x = 0

x^{\frac{3}{2}} - 2 x^{\frac{3}{4}} + 1 = 0

u - 8 = 2

x = - 2

What is the answer to cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4 ?
The answer to cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4 is x=2,x=-2
Solve for x: cube root of 2+x}+\sqrt[3]{2-x}=\sqrt[3]{4
The solution is x=2,x=-2