Nice Algebra Challenge from Olympiads | Can You Crack It?

[@SidneiMV]

x + 1/x = a x² = ax - 1 x³ = ax² - x = a(ax - 1) - x x³ = (a² - 1)x - a x⁴ = (a² - 1)x² - ax = (a² - 1)(ax - 1) - ax x⁴ = (a³ - 2a)x - (a² - 1) x⁵ = (a³ - 2a)x² - (a² - 1)x x⁵ = (a³ - 2a)(ax - 1) - (a² - 1)x x⁵ = (a⁴ - 3a² + 1)x - (a³ - 2a) x⁶ = .....

[@lukaskamin755]

Do you have a video on defining if bigger numbers (say 3-digit numbers are prime)? should I check all the known rules as for if it divisable by all simple numbers smaller than a squzre root of that number? Of course don't mean using internet or any apps for that reason, I mean if it's an exam or an olympiad, where you can't use computers for that

[@SidneiMV]

x⁷ + 1/x⁷ = 843 find x³ + 1/x³ + x⁵ + 1/x⁵ x + 1/x = a x² + 1/x² = a² - 2 (x² + 1/x²)(x + 1/x) = x³ + 1/x³ + x + 1/x x³ + 1/x³ = a(a² - 2) - a x³ + 1/x³ = a³ - 3a (x³ + 1/x³)(x + 1/x) = x⁴ + 1/x⁴ + x² + 1/x² x⁴ + 1/x⁴ = a(a³ - 3a) - (a² - 2) x⁴ + 1/x⁴ = a⁴ - 4a² + 2 (x⁴ + 1/x⁴)(x + 1/x) = x⁵ + 1/x⁵ + x³ + 1/x³ (x⁴ + 1/x⁴)(x³ + 1/x³) = x⁷ + 1/x⁷ + x + 1/x ..... I don't know how to continue..

[@Fjfurufjdfjd]

141

[@kassuskassus6263]

Let a=x+1/x and calcualte x^7+1/x^7. We find a^7-7a^5+14a^3-7a and this is equal to 843. Reolve that for a. We get a=3. Since x^3+1/x^3=a^3-3a, it is equal to 18 and x^5+1/x^5 =123, then x^3+x^5+1/x^3+1/x^5=18+123=141.

[@RashmiRay-c1y]

Let x+1/x=t. Then, x^3+1/x^3 = t^3-3t and x^5+1/x^5 = t^5-5t^3+5t and x^7+1/x^7 = t^7-7t^5+14t^3-7t = 343. t=3 solves this. Thus, the expression we have to evaluate, x^5+1/x^5 + x^3+1/x^3 = t^5-4t^3+2t = 141.

[@mohammedsaysrashid3587]

Thank you Sir 🙏 ...for sharing

[@abcekkdo3749]

141