Question
Solve the cube root equation
\sqrt[3]{x+158} - \sqrt[3]{x-158} = 4
Answer
Let
a = \sqrt[3]{x+158}
(1)
b = \sqrt[3]{x-158}
(2)
Then
a - b = 4
(3)
Difference of a and b raised to the power of 3 gives
a^3 - b^3 = 316
(4)
Factor LHS using difference of cubes formula
(a - b)(a^2+ab+b^2) = 316
(5)
Substitute (3) to (5) and simplify
a^2+ab+b^2 = 79
(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 158}{4} - 4^2 ) = 21
(8)
Substituting ab to (6) gives
a^2+b^2 = 79 - ab = 58
(9)
Construct a perfect square binomial with (9) and (8)
(a+b)^2 = a^2+b^2+2ab
(a+b)^2 = 58 +2\cdot 21 = 100
Taking square root of both sides gives
a +b = \pm10
(10)
Now we get two systems of equations
Case 1
\begin{cases} a+b = 10 \\ a - b =4 \end{cases}
which gives
\begin{cases} a= 7 \\ b =8 \end{cases}
Then
x = a^3 - 158 = 7^3 - 158 = 185
Case 2
\begin{cases} a+b = -10 \\ a - b =4 \end{cases}
which gives
\begin{cases} a= -3 \\ b =-7 \end{cases}
Then
x = a^3 - 158 = -3^3 - 158 = -185
In summary, we get two solutions for x.
x = 185 \text{ or }x = -185