Or... observe that a root may be such that x^3 is a small integer less than 3. 1 doesn't work. 2 does. So first root is 2. If a second real root exists it must be less than 1 and greater than zero due to the general shape of the curve resulting from the radical. However, the quantity inside the radical is greater than 704 for x less than 1. And the root is less than 1 making it a power. So, with declining x less than 1, the LHS increases. Therefore, there is no second real root. The other two roots have to be imaginary. And since the solution is specified as real, we need not solve for them.
LHS as 256^1/8
The given equation is equivalent to 12-x^3 = 2^(x^3-6). This has only one real solution, x=2.
χ=2
x= 2
Or... observe that a root may be such that x^3 is a small integer less than 3. 1 doesn't work. 2 does. So first root is 2. If a second real root exists it must be less than 1 and greater than zero due to the general shape of the curve resulting from the radical. However, the quantity inside the radical is greater than 704 for x less than 1. And the root is less than 1 making it a power. So, with declining x less than 1, the LHS increases. Therefore, there is no second real root. The other two roots have to be imaginary. And since the solution is specified as real, we need not solve for them.
x^3 ➖ 64x^3 m ➖ (12 ➖ x^3)^2,= x^3 ➖ 64x^3(144 ➖ x^9}=(x^3 )^2➖ (64x^3)^2 ➖ 135 {x^9 ➖ 4296x^81}= {4296x^72 ➖ 135}=70.560 ^2^35.10^50^30^2 2^5^7.2^5^5^10^5^6^2.1^1^1.1^1^1^2^5^1^3^2^1 1^1^3^2 32 (x ➖ 3x+2)
2^x³ = 64(12 - x³)
2^(x³ - 6) = 12 - x³
12 - x³ = u => x³ = 12 - u
2^(6 - u) = u
u2ᵘ = 2⁶ [ 2⁶ = (4)2⁴ ]
(uln2)eᵘˡⁿ² = 2⁶ln2
(uln2)eᵘˡⁿ² = 42⁴ln2 = 2⁴ln(2⁴)
W[(uln2)eᵘˡⁿ²] = W[2⁴ln(2⁴)]
uln2 = ln2⁴ = 4ln2
u = 4 => x³ = 8 => *x = 2*
X=2
X=2
X=2