﻿ Solve the cube root equation \sqrt[3]{x+32} - \sqrt[3]{x-32} = 4

#### Question

Solve the cube root equation

\sqrt[3]{x+32} - \sqrt[3]{x-32} = 4

Collected in the board: Cube root

Steven Zheng posted 5 days ago

Let

a = \sqrt[3]{x+32}
(1)
b = \sqrt[3]{x-32}
(2)

Then

a - b = 4
(3)

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

a^3 - b^3 = 64
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 64
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 16
(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 32}{4} - 4^2 ) = 0
(8)

Substituting ab to (6) gives

a^2+b^2 = 16 - ab = 16
(9)

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

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 16 +2\cdot 0 = 16

Taking square root of both sides gives

a +b = \pm4
(10)

Now we get two systems of equations

Case 1

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

which gives

\begin{cases} a= 4 \\ b =2 \end{cases}

Then

x = a^3 - 32 = 4^3 - 32 = 32

Case 2

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

which gives

\begin{cases} a= 0 \\ b =-4 \end{cases}

Then

x = a^3 - 32 = 0^3 - 32 = -32

In summary, we get two solutions for x.

x = 32 \text{ or }x = -32

Steven Zheng posted 5 days ago

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