Solve the cube root equation
$$\sqrt[3]{x+148} - \sqrt[3]{x-148} = 2 $$

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Uniteasy Admin · Accepted Answer

Let			
	$$a = \sqrt[3]{x+148} $$nc		
	$$b = \sqrt[3]{x-148} $$nc		
Then			
	$$a - b = 2$$nc		
Difference of $$a$$ and $$b$$ raised to the power of 3 gives			
	$$a^3 - b^3 = 296$$nc		
Factor LHS using difference of cubes formula			
	$$(a - b)(a^2+ab+b^2) = 296$$nc		
Substitute (3) to (5) and simplify 			
	$$a^2+ab+b^2 = 148$$nc		
Square of equation (3) gives			
	$$(a -b )^2 = 4$$c		
Expand LHS			
	$$a^2 - 2ab +b^2 = 4$$nc		
Substract (7) from (6) and solve for $$ab$$			
	$$ ab = \dfrac{1}{3}(\dfrac{2\cdot 148}{2} - 2^2 ) = 48$$nc		
Substituting $$ab$$ to (6) gives			
	$$a^2+b^2 = 148 - ab = 100$$nc		
Construct a perfect square binomial with (9) and (8)			
	$$(a+b)^2 = a^2+b^2+2ab$$c		
	$$(a+b)^2 = 100 +2\cdot 48 = 196$$c		
Taking square root of both sides gives			
	$$a +b = \pm14$$nc		
Now we get two systems of equations			
Case 1			
	$$\begin{cases} a+b = 14 \ a - b =2 \end{cases}$$c		
which gives			
	$$\begin{cases} a= 8 \ b =13 \end{cases}$$c		
Then			
	$$x = a^3 - 148 = 8^3 - 148 = 364$$c		
Case 2			
	$$\begin{cases} a+b = -14 \ a - b =2 \end{cases}$$c		
which gives			
	$$\begin{cases} a= -6 \ b =-8 \end{cases}$$c		
Then			
	$$x = a^3 - 148 = -6^3 - 148 = -364$$c		
In summary, we get two solutions for $$x$$.			
	$$x = 364 	ext{ or }x = -364$$c

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