Substitute 1/x=y-6 or x=1/(y-6) into the given equation and rearrange to (y+1)^4-(y-1)^4-240=2*4(y^3+y)-240=0 Thus, y^3+y-30=0. By inspection y=3. Divide by (y-3): y^2+3y+10=0, which has roots y=(-3±i√31)/2 From x=1/(y-6), x=1/(3-6)=-1/3 and x=2/(-15±i√31)=2(15±i√31)/194=(15±i√31)/97
Substitute 1/x=y-6 or x=1/(y-6) into the given equation and rearrange to (y+1)^4-(y-1)^4-240=2*4(y^3+y)-240=0
Thus, y^3+y-30=0. By inspection y=3. Divide by (y-3): y^2+3y+10=0, which has roots y=(-3±i√31)/2
From x=1/(y-6), x=1/(3-6)=-1/3 and x=2/(-15±i√31)=2(15±i√31)/194=(15±i√31)/97
Otimo problema
240 = 256-16 = (4^4). -( 2^4). Therefore
5+(1/x) = 2
x = -1/3