Solve the cube root equation
$$\sqrt[3]{x+244} - \sqrt[3]{x-244} = 2 $$

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Uniteasy Admin · Accepted Answer

Let		
	$$a = \sqrt[3]{x+244} $$nc	
	$$b = \sqrt[3]{x-244} $$nc	
Then		
	$$a - b = 2$$nc	
Difference of $$a$$ and $$b$$ raised to the power of 3 gives		
	$$a^3 - b^3 = 488$$nc	
Factor LHS using difference of cubes formula		
	$$(a - b)(a^2+ab+b^2) = 488$$nc	
Substitute (3) to (5) and simplify 		
	$$a^2+ab+b^2 = 244$$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 $$ab$$		
	$$ ab = \dfrac{1}{3}(\dfrac{2\cdot 244}{2} - 2^2 ) = 80$$nc	
Substituting $$ab$$ to (6) gives		
	$$a^2+b^2 = 244 - ab = 164$$nc	
Construct a perfect square binomial with (9) and (8)		
	$$(a+b)^2 = a^2+b^2+2ab$$c	
	$$(a+b)^2 = 164 +2\cdot 80 = 324$$c	
Taking square root of both sides gives		
	$$a +b = \pm18$$nc	
Now we get two systems of equations		
Case 1		
	$$\begin{cases} a+b = 18 \ a - b =2 \end{cases}$$c	
which gives		
	$$\begin{cases} a= 10 \ b =17 \end{cases}$$c	
Then		
	$$x = a^3 - 244 = 10^3 - 244 = 756$$c	
Case 2		
	$$\begin{cases} a+b = -18 \ a - b =2 \end{cases}$$c	
which gives		
	$$\begin{cases} a= -8 \ b =-10 \end{cases}$$c	
Then		
	$$x = a^3 - 244 = -8^3 - 244 = -756$$c	
In summary, we get two solutions for $$x$$.		
	$$x = 756 	ext{ or }x = -756$$c

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