Solve the cube root equation

\sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4

Collected in the board: Algebraic equation

Steven Zheng posted 1 year ago

Answer 1

The root of cubic equation z^3+pz+q=0 is

z=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}

The equation \sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4 is identical to the root equation of the cubic equation. To find x is to determine the value of the expression \dfrac{q^2}{4}+\dfrac{p^3}{27}

Comparing the two equations gives,

-\dfrac{q}{2}=14 \to q=-28 Plug in the values of q and x to the cubic equation

4^3+4p-28=0 \to p=-9


x = \dfrac{q^2}{4}+\dfrac{p^3}{27} = \dfrac{28^2}{4}+\dfrac{(-9)^3}{27}=169

Steven Zheng posted 1 year ago

Answer 2

Evaluating the cube root equation

\sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4


a =\sqrt[3]{14+\sqrt{x} }
b=\sqrt[3]{14-\sqrt{x} }



Raise a and b to their 3rd power and add the equations


Multiplying a by b yields the product

ab = \sqrt[3]{14+\sqrt{x} } \cdotp \sqrt[3]{14-\sqrt{x} } =\sqrt[3]{14^2-x}

Raise both sides of the equation to the 3rd power and obtain


Using the formula for a cube of a binomial

(a+b)^3 = a^3+b^3+3ab(a+b)

Substitution of (3) and (4) yields

4^3 = 28+3\cdotp 4ab

Solve the equation, we get

ab = 3

Plug in the result of ab to (5), we obtain

x=196-3^3 =169

Steven Zheng posted 1 year ago

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