Question

Solve the cube root equation

\sqrt[3]{x+224} - \sqrt[3]{x-224} = 4

Collected in the board: Cube root

Steven Zheng posted 5 days ago

Answer

Let

a = \sqrt[3]{x+224}
(1)
b = \sqrt[3]{x-224}
(2)

Then

a - b = 4
(3)

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

a^3 - b^3 = 448
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 448
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 112
(6)

Square of equation (3) gives

(a -b )^2 = 16

Expand LHS

a^2 - 2ab +b^2 = 16
(7)

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

ab = \dfrac{1}{3}(\dfrac{2\cdot 224}{4} - 4^2 ) = 32
(8)

Substituting ab to (6) gives

a^2+b^2 = 112 - ab = 80
(9)

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

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 80 +2\cdot 32 = 144

Taking square root of both sides gives

a +b = \pm12
(10)

Now we get two systems of equations

Case 1

\begin{cases} a+b = 12 \\ a - b =4 \end{cases}

which gives

\begin{cases} a= 8 \\ b =10 \end{cases}

Then

x = a^3 - 224 = 8^3 - 224 = 288

Case 2

\begin{cases} a+b = -12 \\ a - b =4 \end{cases}

which gives

\begin{cases} a= -4 \\ b =-8 \end{cases}

Then

x = a^3 - 224 = -4^3 - 224 = -288

In summary, we get two solutions for x.

x = 288 \text{ or }x = -288


Steven Zheng posted 5 days ago

Scroll to Top