A Nice Math Olympiad Algebra Problem | How to solve?

[@wes9627]

Substitute x=y-5 into the given equation and rearrange to (y+1)^4-(y-1)^4-80=2*4(y^3+y)-80=0. Divide by 8: y^3+y-10=0. By inspection y=2. Divide by (y-2): y^2+2y+5=0, which has roots y=(-2±4i)/2=-1±2i. Thus, x=y-5=2-5=-3 and -1±2i -5=-6±2i

[@SALogics]

Very nice! ❤

[@denisrenaldo3506]

I don’t understand why using m=x+5. If you want to be shorter (and more magic), it’s better to use m=x+3. You obtain m(m^2+6m+13)=0. So, m=0 or m=-3+2i or m=-3-2i. Finally, x=-3 or x=-6+2i or x=-6-2i.

[@SALogics]

Very nice! ❤

[@zawatsky]

4:00 [(a+b)+(a-b)][(a+b)-(a-b)]=(a+b+a-b)(a+b-a+b)=2a*2b=4a²b². Так сосчитать гораздо проще.

[@SALogics]

Very nice! ❤

[@saaawa]

{(x + 6)^2 + (x + 4)^2}{(x + 6)^2 - (x + 4)^2} = 80 (x^2 + 10x + 26)(x + 5) = 10 x^3 + 15x^2 + 76x + 120 = 0 (x + 3)(x^2 + 12x + 40) = 0 Since from determinant x^2 + 12x + 40 > 0 for any real x, x = -3

[@SALogics]

Very nice! ❤

[@DonRedmond-jk6hj]

Justr using the difference of squares identity to start with will get you to the same place without having to make all the changes of varobles..

[@SALogics]

Very nice! ❤

[@НеллиПшено]

Решаем методом устного счета. 80=81-1 81=3^4 x+6=3 x=-3 Good luck!

[@SALogics]

Very nice! ❤

[@walterwen2975]

Math Olympiad Algebra Problem: (x + 6)⁴ - (x + 4)⁴ = 80 Let: y = x + 5; (x + 6)⁴ - (x + 4)⁴ = (y + 1)⁴ - (y - 1)⁴ = 80 [(y + 1)² - (y - 1)²][(y + 1)² + (y - 1)²] = (4y)[2(y² + 1)] = 80, y³ + y = 10 (y³ - 2³) + (y - 2) = 0, (y - 2)(y² + 2y + 4) + (y - 2) = (y - 2)(y² + 2y + 5) = 0 y - 2 = 0, y = 2 or y² + 2y + 5 = 0, (y + 1)² = - 4 = (2i)², y = - 1 ± 2i y = x + 5, x = y - 5 = 2 - 5 = - 3 or x = (- 1 ± 2i) - 5 = - 6 ± 2i Answer check: x = - 3: (x + 6)⁴ - (x + 4)⁴ = (- 3 + 6)⁴ - (- 3 + 4)⁴ = 3⁴ - 1 = 80; Confirmed x = - 6 ± 2i: (x + 6)⁴ - (x + 4)⁴ = (± 2i)⁴ - (- 2 ± 2i)⁴ = 16i⁴ - 16(- 1 ± i)⁴ (- 1 ± i)² = 1 -/+ 2i + i² = -/+ 2i, (- 1 ± i)⁴ = [(- 1 ± i)²]² = (-/+ 2i)² = 4i² = - 4 (x + 6)⁴ - (x + 4)⁴ = 16i⁴ - 16(- 1 ± i)⁴ = 16 - 16(- 4) = 80; Confirmed Final answer: x = - 3; Two complex value roots, if acceptable; x = - 6 + 2i or x = - 6 - 2i

[@SALogics]

Very nice! ❤