Math Olympiad | Nice Cube Root Radical Equation

Substitution | Sum of Cubes | Factorization | Binomial Theorem | Simultaneous Equations | Cubic Equations | Quadratic Equations | Perfect Square Identity | Cross Method

Chapters:
00:00 Introduction
00:42 Substitution
01:12 Simultaneous Equations
02:18 Quadratic Equations in 2 unknowns a & b
03:09 Quadratic Equation in 1 unknown a

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[mathusiasm]

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[rorydaulton]

I did this another way. Define u as 3x-15, then your equation becomes (where cbrt means cube root)

cbrt(35-u)+cbrt(u)=5.

Change this to cbrt(35-u)=cbrt(u)-5, cube both sides, and simplify, and we get

15[cbrt(u^2)-5cbrt(u)+6]=0. This factors to

15(cbrt(u)-2)(cbrt(u)-3)=0.
cbrt(u)=2 or cbrt(u)=3
u=8 or u=27
3x-15=8 or 3x-15=27
x=23/3 or x=14.