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

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

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

x = 1, x = - 1

Solution steps

Rewrite the equation with

and

x = u^{3} - 1

Solve

Substitute back

solve for

x

Solve

x = 1, x = - 1

Verify Solutions:

x = 1

True

, x = - 1

True

The solutions are

x = 1, x = - 1

2 x + 7 = 5

x - 13 + x + 2 = 5

6 u^{- \frac{1}{2}} + 13 u^{- \frac{1}{4}} + 6 = 0

17 - x = x - 5

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