The equation is equivalent to [(4x+3)/6(3x+4)]³+[2(3x+4)/(4x+3)]³=6 (1/216)•[(4x+3)/(3x+4)]³+2³•[(3x+4)/(4x+3)]³ = 6 or (1/216)•A+8•(1/A)=6 (1), A =[(4x+3)/(3x+4)]³ with x ≠ -4/3, -3/4. (*). Then the (1) A²-1296A+1728=0 => A = 648±264√6=8(81±33√6) (**) From (*) => [(4x+3)/(3x+4)]³ =648±264√6 (4x+3)/(3x+4)=³√(648±264√6)= 6±2√6 => (4x+3)/(3x+4)=6±2√6 => x = [3±14√(3/2)]/10 (**) u³ = 81+33√6 and v³ = 81-33√6 => u³+v³ = 162 and u•v = 3 => etc ..
Let [4x+3]/[18x+24] = a and [6x+8]/[4x+3]=b. Then, a^3+b^3=6 and ab= 1/3. Let a+b=t. Then, 6 = t[t^2-1]. t =2 is the only real root. Thus, a+b=2 and ab=1/3. So, b+1/3b=2 and 3b^2-6b+1=0 which gives b= 1/6[6+/-√(36-12)]=1+/-√6/3. Thus, x= [3√6 -15]/[6-4√6], -[3√6+15]/[6+4√6].
The equation is equivalent to
[(4x+3)/6(3x+4)]³+[2(3x+4)/(4x+3)]³=6
(1/216)•[(4x+3)/(3x+4)]³+2³•[(3x+4)/(4x+3)]³ = 6 or (1/216)•A+8•(1/A)=6 (1),
A =[(4x+3)/(3x+4)]³ with x ≠ -4/3, -3/4. (*).
Then the (1) A²-1296A+1728=0
=> A = 648±264√6=8(81±33√6) (**)
From (*)
=> [(4x+3)/(3x+4)]³ =648±264√6
(4x+3)/(3x+4)=³√(648±264√6)=
6±2√6 =>
(4x+3)/(3x+4)=6±2√6 =>
x = [3±14√(3/2)]/10
(**) u³ = 81+33√6 and v³ = 81-33√6 =>
u³+v³ = 162 and u•v = 3 => etc ..
Let [4x+3]/[18x+24] = a and [6x+8]/[4x+3]=b. Then, a^3+b^3=6 and ab= 1/3. Let a+b=t. Then, 6 = t[t^2-1]. t =2 is the only real root. Thus, a+b=2 and ab=1/3. So, b+1/3b=2 and 3b^2-6b+1=0 which gives b= 1/6[6+/-√(36-12)]=1+/-√6/3. Thus, x= [3√6 -15]/[6-4√6], -[3√6+15]/[6+4√6].
Все гораздо проще
18х+24=3*(6х+8)
Тогда
((4х+3)/(6х+8))^3=у
Дальнейшее просто
у/27+1/у=6
Обычное квадратное уравнение
Θετω α=(4χ+3)/(3χ+4) χ=/-3/4 ,-4/3
Αν α/6+2/α=ψ τοτε ψ^3-ψ-6=0
(ψ-2)(ψ^2+2ψ+3)=0 ψ=2 .. .(Δ