Question

Solve the cube root equation

\sqrt[3]{x+62} - \sqrt[3]{x-62} = 4

Collected in the board: Cube root

Steven Zheng posted 5 days ago

Answer

Let

a = \sqrt[3]{x+62}
(1)
b = \sqrt[3]{x-62}
(2)

Then

a - b = 4
(3)

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

a^3 - b^3 = 124
(4)

Factor LHS using difference of cubes formula

(a - b)(a^2+ab+b^2) = 124
(5)

Substitute (3) to (5) and simplify

a^2+ab+b^2 = 31
(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 62}{4} - 4^2 ) = 5
(8)

Substituting ab to (6) gives

a^2+b^2 = 31 - ab = 26
(9)

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

(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 26 +2\cdot 5 = 36

Taking square root of both sides gives

a +b = \pm6
(10)

Now we get two systems of equations

Case 1

\begin{cases} a+b = 6 \\ a - b =4 \end{cases}

which gives

\begin{cases} a= 5 \\ b =4 \end{cases}

Then

x = a^3 - 62 = 5^3 - 62 = 63

Case 2

\begin{cases} a+b = -6 \\ a - b =4 \end{cases}

which gives

\begin{cases} a= -1 \\ b =-5 \end{cases}

Then

x = a^3 - 62 = -1^3 - 62 = -63

In summary, we get two solutions for x.

x = 63 \text{ or }x = -63


Steven Zheng posted 5 days ago

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