What is the answer to \sqrt[3]{x^3-5x-4}=x-1 ?

The answer to \sqrt[3]{x^3-5x-4}=x-1 is x=-1/3 ,x=3

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\sqrt[3]{x^3-5x-4}=x-1

x = - \frac{1}{3}, x = 3

Decimal Notation

x = - 0.33333 \dots, x = 3

Solution steps

Take both sides of the equation to the power of

3 : x^{3} - 5 x - 4 = x^{3} - 3 x^{2} + 3 x - 1

x^{3} - 5 x - 4 = x^{3} - 3 x^{2} + 3 x - 1

Solve

x^{3} - 5 x - 4 = x^{3} - 3 x^{2} + 3 x - 1 : x = - \frac{1}{3}, x = 3

x = - \frac{1}{3}, x = 3

Verify Solutions:

x = - \frac{1}{3}

True

, x = 3

True

The solutions are

x = - \frac{1}{3}, x = 3

2^{5} (x + 6)^{3} + 3 = 19

1 = 2 π \frac{x}{9.81}

x + 2 - x = 2

x + 1 - x - 2 = 1

x, x^{\frac{4}{3}} = 16

What is the answer to \sqrt[3]{x^3-5x-4}=x-1 ?
The answer to \sqrt[3]{x^3-5x-4}=x-1 is x=-1/3 ,x=3
Solve for x: \sqrt[3]{x^3-5x-4}=x-1
The solution is x=-1/3 ,x=3