What is the answer to solvefor x,3y=\sqrt[3]{x+2}+5 ?

The answer to solvefor x,3y=\sqrt[3]{x+2}+5 is x=27y^3-135y^2+225y-127

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solvefor x,3y=\sqrt[3]{x+2}+5

solve for

x = 27 y^{3} - 135 y^{2} + 225 y - 127

Solution steps

Switch sides

Move

5

to the right side

Take both sides of the equation to the power of

3 : x + 2 = 27 y^{3} - 135 y^{2} + 225 y - 125

x + 2 = 27 y^{3} - 135 y^{2} + 225 y - 125

Solve

x + 2 = 27 y^{3} - 135 y^{2} + 225 y - 125 : x = 27 y^{3} - 135 y^{2} + 225 y - 127

x = 27 y^{3} - 135 y^{2} + 225 y - 127

Verify Solutions:

x = 27 y^{3} - 135 y^{2} + 225 y - 127

True

The solution is

x = 27 y^{3} - 135 y^{2} + 225 y - 127

4 = 7 - 2 25 + 18 x

k^{\frac{- 1}{2}} = 0.04

a = x^{2} - 4

4 x - 7 - 5 = 0

What is the answer to solvefor x,3y=\sqrt[3]{x+2}+5 ?
The answer to solvefor x,3y=\sqrt[3]{x+2}+5 is x=27y^3-135y^2+225y-127