Math Olympiad | A Nice Algebra Math Problem | Find the Value of x

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@PradeepShrotriya-wi2me]

Nice video

[@RyanLewis-Johnson-wq6xs]

1/(x^3)-1/(x+1)^3=7/8 x=-2 x=1 x=(-1/2)±(1/14)iSqrt[147±28Sqrt[21] ]=-0.5±0.1(0.714285 recurring)iSqrt[147±28Sqrt[21]]

[@shashankkatiha9439]

For solving {1/x^3 - 1/(x+1)^3}=7/8, I resolved it to (1/x - 1/(x+1))(1/x^2 + 1/x(x + 1) + 1/(x +1)^2)= 7/8 as I used the formula for a^3 - b^3. {1/x(x +1)}{1/x^2 + 1/(x+1)^2 + 1/x(x +1)} = 7/8.{1/x(x +1)}{(1/x - 1/(x+1))^2 + 2/x(x + 1) + 1/x(x + 1)} = 7/8 using a^2 + b^2 = (a - b)^2 + 2ab. {1/x(x + 1)}{(1/x(x + 1))^2 + 3/x(x + 1)} = 7/8 Putting 1/x(x + 1) = a. a(a^2 + 3a) = 7/8. a^3 + 3a^2 = 7/8. 8a^3 + 24a^2 = 7. bringing everything to lhs 8a^3 + 24a^2 - 7 = 0. Here if we put a = 1/2, 8(1/2)^3 + 24(1/2)^2 - 7 = 8(1/8) + 24(1/4) - 7 = 1 + 6 - 7 = 0. So we divide by a - 1/2 using synthetic division. This gave (a - 1/2)(4a^2 + 4a + 17) = 0 So a = 1/2. Other roots are (-7 + sqrt(21))/4,(-7 - sqrt(21))/4. Putting back 1/x(x + 1) = 1/2. x(x + 1) = 2 x^2 + x = 2 x^2 + x - 2 = 0 x^2 + 2x - x - 2 = 0 x(x + 2) -1(x +2) = 0 (x - 1)(x + 2) = 0 x = 1,-2

[@raghvendrasingh1289]

(x+1)^3 - x^3 = 7{ (x^2+x)^3 }/8 8 {3(x^2+x) +1} -7 (x^2+x)^3 = 0 7t^3 -24t- 8 = 0 where t = x^2+x 7t^3 -28t +4t-8 = 0 7t (t+2)((t-2) +4 (t-2) = 0 (t-2) (7t^2+14t+4) =0 now use x^2+x= t