Germany l A Nice Algebra Problem l Math Olympiad

[@itzmrinyy7484]

Let x/3 = u u⁶-2⁶=0 Difference of squares: (u³-2³)(u³+2³) Sum and difference of cubes: (u+2)(u²-2u+4)(u-2)(u²+2u+4)=0 Via quadratic formula: u=±1±i√3, ±2 Input u=x/3: x = ±6, ±3 ± 3i√3

[@key_board_x]

(x/3)⁶ = 64 x⁶/3⁶ = 2⁶ x⁶ = 3⁶ * 2⁶ x⁶ = 6⁶ ← this is a complex number (particular) The modulus of x⁶ is 6⁶, so the modulus of x is 6. The argument of x⁶ is 0, so the argument of x is 0/6, i.e. 0. Then, the first root of x⁶ is: x1 = 6.[cos(0) + i.sin(0)] → you add angle of (2π/6), i.e. (π/3) to get the second root x2 = 6.[cos(0 + {π/3}) + isin(0 + {π/3})] → you add angle of (π/3) again to get the third root x3 = 6.[cos(0 + {π/3} + {π/3}) + isin(0 + {π/3} + {π/3})] → you add angle of (π/3) once again to get the fourth root x4 = 6.[cos(0 + {π/3} + {π/3} + {π/3}) + isin(0 + {π/3} + {π/3} + {π/3})] → again to get the fifth root x5 = 6.[cos(0 + {π/3} + {π/3} + {π/3} + {π/3}) + isin(0 + {π/3} + {π/3} + {π/3} + {π/3})] → and a last time to get the last one x6 = 6.[cos(0 + {π/3} + {π/3} + {π/3} + {π/3} + {π/3}) + isin(0 + {π/3} + {π/3} + {π/3} + {π/3} + {π/3})] It gives x1 = 6.[cos(0) + i.sin(0)] x1 = 6.[1 + 0] x1 = 6 x2 = 6.[cos(π/3) + i.sin(π/3)] x2 = 6.[(1/2) + i.{(√3)/2}] x2 = 3 + 3i√3 x2 = 3.(1 + i√3) x3 = 6.[cos(2π/3) + i.sin(2π/3)] x3 = 6.[(- 1/2) + i.{(√3)/2}] x3 = - 3 + 3i√3 x3 = - 3.(1 - i√3) x4 = 6.[cos(3π/3) + i.sin(3π/3)] x4 = 6.[cos(π) + i.sin(π)] x4 = 6.[- 1 + 0] x4 = - 6 x5 = 6.[cos(4π/3) + i.sin(4π/3)] x5 = 6.[(- 1/2) + i.{- (√3)/2}] x5 = - 3 - 3i√3 x5 = - 3.(1 + i√3) x6 = 6.[cos(5π/3) + i.sin(5π/3)] x6 = 6.[(1/2) + i.{- (√3)/2}] x6 = 3 - 3i√3 x6 = 3.(1 - i√3) Summarize x1 = 6 x2 = 3.(1 + i√3) x3 = - 3.(1 - i√3) x4 = - 6 x5 = - 3.(1 + i√3) x6 = 3.(1 - i√3)

[@RealQinnMalloryu4]

x^6/729=1 (x ➖ 1x+1).

[@gopagon2114]

X=6

[@walterwen2975]

Germany, Math Olympiad: (x/3)⁶ = 64; x =? First method: (x/3)⁶ = 64 = 8² = 2⁶ = (± 6/3)⁶; x = ± 6, Missing 4 complex value roots Second method: Solve the equation directly; Let: y = x/3; y⁶ = 64, y⁶ - 2⁶ = 0, (y³ - 2³)(y³ + 2³) = 0 [(y - 2)(y² + 2y + 4)][(y + 2)(y² - 2y + 4)] = 0 y - 2 = 0; y + 2 = 0; y² + 2y + 4 = 0, (y + 1)² = - 3 or y² - 2y + 4 = 0, (y - 1)² = - 3 y = 2, x = 6; y = - 2, x = - 6; y = - 1 ± i√3, x = - 3 ± 3i√3; y = 1 ± i√3, x = 3 ± 3i√3 Answer check: x = ± 6: (x/3)⁶ = 64; Confirmed as shown in First method x = - 3 ± 3i√3, y = x/3: y² + 2y + 4 = 0, y² = - 2(y + 2) y⁴ = [- 2(y + 2)]² = 4(y² + 4y + 4) = 4(y² + 2y + 4 + 2y) = 4(2y) = 8y y⁶ = (y²)(y⁴) = [- 2(y + 2)](8y) = - 16(y² + 2y) = - 16(- 4) = 64; Confirmed x = 3 ± 3i√3, y = x/3: y² - 2y + 4 = 0, y² = 2(y - 2) y⁴ = [2(y - 2)]² = 4(y² - 4y + 4) = 4(y² - 2y + 4 - 2y) = 4(- 2y) = - 8y y⁶ = (y²)(y⁴) = [2(y - 2)](- 8y) = - 16(y² - 2y) = - 16(- 4) = 64; Confirmed Final answer: x = 6; x = - 6; x = - 3 + 3i√3; x = - 3 - 3i√3; x = 3 + 3i√3 or x = - 3 - 3i√3

[@raghvendrasingh1289]

👍 Method 1 Let x = 3 t (t^3)^2 - 8°2 = 0 (t^3+8)(t^3 - 8) = 0 (t+2)(t - 2)(t^2 - 2 t+4)(t^2+2 t+4) = 0 (t+2)(t - 2) { (t - 1)^2 - (i√3)^2 } { (t+1)^2 - (i√3^2 } = 0 there are 6 values of t , now x = 3 t hence x = - 6 , 6 , 3+3√3 i , 3 - 3√3 i , - 3+3√3 I , - 3 - 3√3 i Method 2 x^6 = 6^6 x^6 = 6^6 (cos 2 r π +i sin 2 r π) , r is an integer by Demoivre's theorem x = 6 (cos r π/3) + i sin r π/3) r = 0 , 1 , 2 , 3 , 4 , 5 now we can find 6 roots

[@vijaymaths5483]

Great job! That's a very creative approach to solving the problem.

[@prollysine]

x^6+/-x^5+/-x^4+/-x^3+/-x^2+/-x-46656=0 , (x-6)(x^5+6x^4+36x^3+216x^2+1296x+7776)=0 , (x+6)(x^4+36x^2+1296)=0 , 1 -6 1 6 36 216 1296 7776 6 -36 ---------- ------------------ --------------------- 36 -216 solu , x= 6 , -6 , -3+iV3 , -3+iV3 , 3+iV3 , 3-iV3 , 216 -1296 1296 -7776 7776 -46656