x = 1/7 or x = -11/8 Let T = (11+8x)/(2+3x) and S = (1-7x)/(2+3x) , x ≠ -2/3.(*) Then T - S = 5 (1) and T³ - S³ =125 (2) Solving the system of (1), (2) T = 5 and S = 0 => x = 1/7 or x = -11/8. Another approach. The given equation is equivalent to (11+8x)³ -(1-7x)³=125(2+3x)³ (11+8x)³ +(7x-1)³ +(-5(2+3x))³ = 0. (1) Let a=11+8x, b=7x-1, c=-10-15x. Then a+b+c=0. From known identity => a³+b³+c³ =3abc (2) The (2) due to (2) rewrite as 0 =3(11+8x)(7x-1)(-10-15x) => x =-11/8 or x=1/7 or x=-2/3. x=-2/3 not accepted due to (*). Another approach. The given equation is equivalent to (11+8x)³-(1-7x)³=125(2+3x)³ ... 168x³+319x²+105x-22=0 => x=-11/8, x=1/7, x=-2/3. The x=-2/3 not accepted due to (*).
((11+8(-1.375))/(2+3(-1.375)))^3-((1-7(-1.375))/(2+3(-1.375)))^3=125 X=-1.375=-11/8=-1 3/8 X=0.142857 recurring=1/7 It’s in my head.
Let a=[11+8x]/[2+3x] and b=[1-7x]/[2+3x]. So, a-b=5 and a^3-b^3=125. Let t=ab. Then 125=5(25+3t) which gives t=0. So, a=0 or b=0. Thus, x=-11/8, 1/7.
x = 1/7 or x = -11/8
Let T = (11+8x)/(2+3x) and
S = (1-7x)/(2+3x) , x ≠ -2/3.(*)
Then T - S = 5 (1) and T³ - S³ =125 (2)
Solving the system of (1), (2)
T = 5 and S = 0 =>
x = 1/7 or x = -11/8.
Another approach.
The given equation is equivalent to
(11+8x)³ -(1-7x)³=125(2+3x)³
(11+8x)³ +(7x-1)³ +(-5(2+3x))³ = 0. (1)
Let a=11+8x, b=7x-1, c=-10-15x.
Then a+b+c=0.
From known identity =>
a³+b³+c³ =3abc (2)
The (2) due to (2) rewrite as
0 =3(11+8x)(7x-1)(-10-15x) =>
x =-11/8 or x=1/7 or x=-2/3.
x=-2/3 not accepted due to (*).
Another approach.
The given equation is equivalent to
(11+8x)³-(1-7x)³=125(2+3x)³
... 168x³+319x²+105x-22=0 =>
x=-11/8, x=1/7, x=-2/3.
The x=-2/3 not accepted due to (*).
Dopo semplici calcoli(differenza di cubi)rimane la quadrica 168x^2+207x-33=0...x=-11/8..x=1/7
X=-11/8; 1/7