Online Course Discussion Forum

Math Challenge III Algebra 3.32

 
 
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Math Challenge III Algebra 3.32
by James Shi - Thursday, 6 August 2020, 11:07 PM
 

For this question, my first idea was to substitute $a = (x + 1)^{\frac{1}{3}}$ and $b = (x - 1)^{\frac{1}{3}}$. I then get the equation $a^2 + 4b^2 = 5ab$. I completed the square and got $(a - 2b)^2 = ab$. I thought I could construct a quadratic with $a$ and $b$ as the roots, but I do not think I can do that very easily. Are there any other strategies that I could try for this problem? Thank you.

 
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Re: Math Challenge III Algebra 3.32
by James Shi - Friday, 7 August 2020, 5:37 PM
 

Nevermind, I cubed both sides of the equation and got the answer.