A difficult olympiad math problem | China Math Olympiad Questions

[@pizza8567]

This is a very straightforward question, you could just plug in 1 and get the answer aswell.

[@MathBeast.channel-l9i]

Alright 👍

[@pizza8567]

@ but your way actually shows how you got it lol

[@thundervolt97]

​@@pizza8567man the degree is 4 so there are 4 solutions to this equation not only 3 so

[@CNunez-vn1hr]

This is a typical four-terms polynomial. To find the solution in two steps, first factor by grouping and then use the zero-factor property to get the solutions. Assuming that you know how to factor the difference of two perfect cubes, you should be able to solve it in one minute. Btw, there is a double real-zero (x=1), and the two complex number zeros; i.e., four solutions or zeros matching the degree 4 of this polynomial. These complex zeros must be written in the standard form: a+bi. Hope this helps.

[@CNunez-vn1hr]

Happy new year!

[@MathBeast.channel-l9i]

Happy New Year Sir 🎈

[@MathBeast.channel-l9i]

Alright Boss Nice Approach 👍

[@curiousityofmind]

very esay question

[@MathBeast.channel-l9i]

Alright Boss

[@walterwen2975]

China Math Olympiad Question: x⁴ - x³ - x + 1 = 0; x =? x⁴ - x³ - x + 1 = (x⁴ - x³) - (x - 1) = x³(x - 1) - (x - 1) = (x - 1)(x³ - 1) = 0 (x - 1)(x - 1)(x² + x + 1) = [(x - 1)²](x² + x + 1) = 0; x - 1 = 0 or x² + x + 1 = 0 x = 1, Double root or x = (- 1 ± i√3)/2 Answer check: x = 1: x⁴ - x³ - x + 1 = 1 - 1 - 1 + 1 = 0; Confirmed x = (- 1 ± i√3)/2: x² + x + 1 = 0, x² = - x - 1, x³ = - x² - x = 1, x⁴ = x x⁴ - x³ - x + 1 = x - 1 - x + 1 = 0; Confirmed Final answer: x = 1, Double root; x = (- 1 + i√3)/2 or x = (- 1 - i√3)/2

[@Kosekans]

Obviously you can split off (x-1) twice. The rest is straightforward.

[@MathBeast.channel-l9i]

Okay 👍

[@prollysine]

x^3(x-1)-(x-1)=0 , (x-1)(x^3-1)=0 , x^3+/-x^2+/-x-1=0 , (x-1)(x^2+x+1)=0 , x^2+x+1=0 , x=(-1+/-V(1-4))/2 , 1 -1 solu , x=1 , (-1+i*V3)/2 , (-1-i*V3)/2 , 1 -1 1 -1

[@RealQinnMalloryu4]

(x^4)^2 ➖ (x^3)^2 ➖ x+1={x^16 ➖ x^9} ➖ x+1 x^7 ➖ (x)^2+1={x^7 ➖ x^2}+1=x^5+{1+1 ➖}={x^5+2 }=2x^5 2x^2^3 1x^2^1 1x^2 (x ➖ 2x+1).