A Nice Mathematics algebra problem | Olympiad Question | radical problem | x=?

[@HuzaifaKhan-m3i]

❤❤❤❤

[@NaeemKhan-q6w1v]

Good

[@HuzaifaKhan-m3i]

Great question sir g

[@AfsarSaid-f8s]

Great question sir

[@gregevgeni1864]

Let t = x+6 , x+4 = t-2, x+6 = t+2 => So (x+6)⁶-16=(x+4)³(x+6)³ => t⁶-16=(t-2)³(t+2)³ => t⁴-4t²+4=0 => t²=2 => t=±√2 => x+6=±√2 => x = -6±√2.

[@SidneiMV]

x + 6 = u ∛(u⁶ - 16) = (u - 2)(u + 2) = u² - 4 u⁶ - 16 = (u² - 4)³ u⁶ - 16 = u⁶ - 64 - 12u²(u² - 4) 12u²(u² - 4) + 48 = 0 u⁴ - 4u² + 4 = 0 (u² - 2)² = 0 u² = 2 => u = ± √2 *x = -6 ± √2*

[@RealQinnMalloryu4]

x^3 (x^6+36) ➖ 16=x^3 36x^6 ➖ (16)^2=x^3{36x^6 256}={x^3* 220x^6}=220x^18 110x^18 10^10^10x^18 10x^2^9 5^5x^2^3^2 2^3^2^3x^1^1^1 1^1^2^3^x 2^3^x (x ➖ 3x+2).(x+4)^6 (x+8)^6 (x^6+24)(x^6+48) ={24x^6+48x^6}=72x^12 24^48x^12 2^2^2^2^2^2^3x^6^6 1^1^1^1^1^1^1^2x^3^3^3^3 1^2x^1^1^1^3 2x^3 (x ➖ 3x+2).