#### Question

Solve the cube root equation

\sqrt[3]{x+4} - \sqrt[3]{x-4} = 2

Collected in the board: Cube root

Steven Zheng posted 5 months ago

Let

a = \sqrt[3]{x+4}
(1)
b = \sqrt[3]{x-4}
(2)

Then

a - b = 2
(3)

Difference of a and b raised to the power of 3 gives

a^3 - b^3 = 8
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 8
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 4
(6)

Square of equation (3) gives

(a -b )^2 = 4

Expand LHS

a^2 - 2ab +b^2 = 4
(7)

Substract (7) from (6) and solve for ab

ab = \dfrac{1}{3}(\dfrac{2\cdot 4}{2} - 2^2 ) = 0
(8)

Substituting ab to (6) gives

a^2+b^2 = 4 - ab = 4
(9)

Construct a perfect square binomial with (9) and (8)

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 4 +2\cdot 0 = 4

Taking square root of both sides gives

a +b = \pm2
(10)

Now we get two systems of equations

Case 1

\begin{cases} a+b = 2 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= 2 \\ b =1 \end{cases}

Then

x = a^3 - 4 = 2^3 - 4 = 4

Case 2

\begin{cases} a+b = -2 \\ a - b =2 \end{cases}

which gives

\begin{cases} a= 0 \\ b =-2 \end{cases}

Then

x = a^3 - 4 = 0^3 - 4 = -4

In summary, we get two solutions for x.

x = 4 \text{ or }x = -4

Steven Zheng posted 5 months ago

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