Solve the cube root equation
$$\sqrt[3]{x+62} - \sqrt[3]{x-62} = 4 $$

Question

Solve the cube root equation	
$$\sqrt[3]{x+62} - \sqrt[3]{x-62} = 4 $$

Uniteasy Admin · Accepted Answer

Let			
	$$a = \sqrt[3]{x+62} $$nc		
	$$b = \sqrt[3]{x-62} $$nc		
Then			
	$$a - b = 4$$nc		
Difference of $$a$$ and $$b$$ raised to the power of 3 gives			
	$$a^3 - b^3 = 124$$nc		
Factor LHS using difference of cubes formula			
	$$(a - b)(a^2+ab+b^2) = 124$$nc		
Substitute (3) to (5) and simplify 			
	$$a^2+ab+b^2 = 31$$nc		
Square of equation (3) gives			
	$$(a -b )^2 = 16$$c		
Expand LHS			
	$$a^2 - 2ab +b^2 = 16$$nc		
Substract (7) from (6) and solve for $$ab$$			
	$$ ab = \dfrac{1}{3}(\dfrac{2\cdot 62}{4} - 4^2 ) = 5$$nc		
Substituting $$ab$$ to (6) gives			
	$$a^2+b^2 = 31 - ab = 26$$nc		
Construct a perfect square binomial with (9) and (8)			
	$$(a+b)^2 = a^2+b^2+2ab$$c		
	$$(a+b)^2 = 26 +2\cdot 5 = 36$$c		
Taking square root of both sides gives			
	$$a +b = \pm6$$nc		
Now we get two systems of equations			
Case 1			
	$$\begin{cases} a+b = 6 \ a - b =4 \end{cases}$$c		
which gives			
	$$\begin{cases} a= 5 \ b =4 \end{cases}$$c		
Then			
	$$x = a^3 - 62 = 5^3 - 62 = 63$$c		
Case 2			
	$$\begin{cases} a+b = -6 \ a - b =4 \end{cases}$$c		
which gives			
	$$\begin{cases} a= -1 \ b =-5 \end{cases}$$c		
Then			
	$$x = a^3 - 62 = -1^3 - 62 = -63$$c		
In summary, we get two solutions for $$x$$.			
	$$x = 63 	ext{ or }x = -63$$c

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