What is the answer to solvefor g,g(\sqrt[3]{216})=x^3-29 ?

The answer to solvefor g,g(\sqrt[3]{216})=x^3-29 is g=(x^3-29)/6

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solvefor g,g(\sqrt[3]{216})=x^3-29

solve for

g, g (3216) = x^{3} - 29

g = \frac{x ^{3} - 29}{6}

Solution steps

g (3216) = x^{3} - 29

Simplify

3216 : 6

g \cdot 6 = x^{3} - 29

Divide both sides by

6

g = \frac{x ^{3} - 29}{6}

x^{3} - 4 x^{2} - 11 x - 6 = 0

x^{3} - 3 x + 5

y, 4 y - 4 x^{3} = 0

2 x^{3} - 18 x^{2} - 44 x = 0

4 x^{5} + 4 x^{3} = 0

What is the answer to solvefor g,g(\sqrt[3]{216})=x^3-29 ?
The answer to solvefor g,g(\sqrt[3]{216})=x^3-29 is g=(x^3-29)/6