Let (x^10-33)^1/5 = y^2 > x^2-y^2=3 and x^10-y^10=33. Let x^2=a and y^2=b. Note that x is positive. So, a-b=3 and a^5-b^5=33. Let ab =t Thus, from (a-b)^5 = a^5-b^5 +10a^2b^2(a-b) -5ab(a^3-b^3), we get t^2+9t+14=0 > t = -2, -7. So, a-b=3 and ab = -2, -7. If ab = -2, we get a=1,2 > x = 1, √2. ab = -7 does not yield a real value for x. So, x=1,√2.
x = √(3 + ⁵√(x¹⁰ -33)) (1) ⁵√(x¹⁰ - 33) = y (2). The (1) due to (2) become x = √(3+y) (3) with y> -3 and x >0 (#). From (3) => x² = 3+y (3)' From (2) => x¹⁰ -33 =y⁵ => (x²)⁵ - y⁵ = 33 (2)'. The (2)' due to (3)' become (3+y)⁵ - y⁵= 33 y⁴ + 6 y³ + 18 y² +27 y + 14=0 (y+1)(y+2)(y² +3y+7) =0 y = -1 or y =-2 (real roots) accepted. So from (3)' : x² = 3-1= 2 => x = ±√2 or x² = 3-2 = 1 => x = ±1 The solutions x = 1 or x =√2, due to (#).
Let (x^10-33)^1/5 = y^2 > x^2-y^2=3 and x^10-y^10=33. Let x^2=a and y^2=b. Note that x is positive. So, a-b=3 and a^5-b^5=33. Let ab =t Thus, from (a-b)^5 = a^5-b^5 +10a^2b^2(a-b) -5ab(a^3-b^3), we get t^2+9t+14=0 > t = -2, -7. So, a-b=3 and ab = -2, -7. If ab = -2, we get a=1,2 > x = 1, √2. ab = -7 does not yield a real value for x. So, x=1,√2.
Sqrt[3+Surd[(x^10-33),5] x=1 x=Sqrt[2] It’s in my head.
X= 1; √2
X=1, (2)^(1/2)
x^10-33=A^5
when x=1, A=-2
So, x=1 is the answer.
{3+3 ➖ }+x^5+x^10 ➖ 33 =6+{x^5+x^5 ➖ } x^10 ➖ 33={6+x^10}+x^10 ➖ 33 =6x^10 (x^10)^2 ➖ (33)^2=6x^10+{x^100 ➖ 1089}={6x^10+989}=995x^10 3^2^3^2^5x2^5 1^11^12^3x^2^2^3.1^3x1^2^1 3x^2 (x ➖ 3x+2).
1 is not a solution. Verification needed
x = √(3 + ⁵√(x¹⁰ -33)) (1)
⁵√(x¹⁰ - 33) = y (2).
The (1) due to (2) become
x = √(3+y) (3)
with y> -3 and x >0 (#).
From (3) => x² = 3+y (3)'
From (2) => x¹⁰ -33 =y⁵ =>
(x²)⁵ - y⁵ = 33 (2)'.
The (2)' due to (3)' become
(3+y)⁵ - y⁵= 33
y⁴ + 6 y³ + 18 y² +27 y + 14=0
(y+1)(y+2)(y² +3y+7) =0
y = -1 or y =-2 (real roots) accepted.
So from (3)' : x² = 3-1= 2 => x = ±√2
or x² = 3-2 = 1 => x = ±1
The solutions x = 1 or x =√2,
due to (#).