Answer :
We are given the equation:
[tex]\[ \sqrt[3]{9 + \sqrt{x}} + \sqrt[3]{9 - \sqrt{x}} = 3 \][/tex]
To find [tex]\( x \)[/tex], let's start by making a substitution:
[tex]\[ a = \sqrt[3]{9 + \sqrt{x}} \quad \text{and} \quad b = \sqrt[3]{9 - \sqrt{x}} \][/tex]
Thus, the equation becomes:
[tex]\[ a + b = 3 \][/tex]
Next, let's cube both sides of the equation to eliminate the cube roots. Recall the binomial expansion:
[tex]\[ (a + b)^3 = a^3 + b^3 + 3ab(a + b) \][/tex]
Since [tex]\( a + b = 3 \)[/tex], we can substitute that into the expanded form:
[tex]\[ 3^3 = a^3 + b^3 + 3ab \cdot 3 \][/tex]
Simplify:
[tex]\[ 27 = a^3 + b^3 + 9ab \][/tex]
We also know:
[tex]\[ a^3 = 9 + \sqrt{x} \quad \text{and} \quad b^3 = 9 - \sqrt{x} \][/tex]
So,
[tex]\[ a^3 + b^3 = (9 + \sqrt{x}) + (9 - \sqrt{x}) = 18 \][/tex]
Substitute [tex]\( a^3 + b^3 = 18 \)[/tex] into the simplified cubic equation:
[tex]\[ 27 = 18 + 9ab \][/tex]
Subtract 18 from both sides:
[tex]\[ 9 = 9ab \][/tex]
Divide both sides by 9:
[tex]\[ 1 = ab \][/tex]
Now, use the fact that [tex]\( ab = 1 \)[/tex] to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. Recall the relationships:
[tex]\[ ab = 1 \][/tex]
[tex]\[ a + b = 3 \][/tex]
We have a system of equations:
1. [tex]\( a + b = 3 \)[/tex]
2. [tex]\( ab = 1 \)[/tex]
These are the roots of the quadratic equation:
[tex]\[ t^2 - (a + b)t + ab = 0 \][/tex]
Replace [tex]\( a + b \)[/tex] and [tex]\( ab \)[/tex]:
[tex]\[ t^2 - 3t + 1 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus,
[tex]\[ a = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad b = \frac{3 - \sqrt{5}}{2} \][/tex]
Remember that:
[tex]\[ a^3 = 9 + \sqrt{x} \quad \text{and} \quad b^3 = 9 - \sqrt{x} \][/tex]
Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] back:
[tex]\[ a^3 = \left( \frac{3 + \sqrt{5}}{2} \right)^3 \][/tex]
[tex]\[ b^3 = \left( \frac{3 - \sqrt{5}}{2} \right)^3 \][/tex]
We now need to compute these cubes:
[tex]\[ \left( \frac{3 + \sqrt{5}}{2} \right)^3 = \frac{(3 + \sqrt{5})^3}{8} \][/tex]
[tex]\[ (3 + \sqrt{5})^3 = 27 + 3 \cdot 9 \cdot \sqrt{5} + 3 \cdot 5 + \sqrt{5}^3 = 27 + 27\sqrt{5} + 15 + 5\sqrt{5} \][/tex]
[tex]\[ = 42 + 32\sqrt{5} \][/tex]
So,
[tex]\[ \left( \frac{3 + \sqrt{5}}{2} \right)^3 = \frac{42 + 32\sqrt{5}}{8} = \frac{21 + 16\sqrt{5}}{4} \][/tex]
Similarly, for [tex]\( b \)[/tex]:
[tex]\[ \left( \frac{3 - \sqrt{5}}{2} \right)^3 = \frac{(3 - \sqrt{5})^3}{8} \][/tex]
[tex]\[ (3 - \sqrt{5})^3 = 27 - 27\sqrt{5} + 15 - 5\sqrt{5} = 42 - 32\sqrt{5} \][/tex]
So,
[tex]\[ \left( \frac{3 - \sqrt{5}}{2} \right)^3 = \frac{42 - 32\sqrt{5}}{8} = \frac{21 - 16\sqrt{5}}{4} \][/tex]
We have:
[tex]\[ 9 + \sqrt{x} = \frac{21 + 16\sqrt{5}}{4} \][/tex]
[tex]\[ 9 - \sqrt{x} = \frac{21 - 16\sqrt{5}}{4} \][/tex]
Subtract these equations:
[tex]\[ (9 + \sqrt{x}) - (9 - \sqrt{x}) = \frac{21 + 16\sqrt{5} - (21 - 16\sqrt{5})}{4} \][/tex]
[tex]\[ 2\sqrt{x} = \frac{32\sqrt{5}}{4} = 8\sqrt{5} \][/tex]
[tex]\[ \sqrt{x} = 4\sqrt{5} \][/tex]
Square both sides to find [tex]\( x \)[/tex]:
[tex]\[ x = (4\sqrt{5})^2 = 16 \times 5 = 80 \][/tex]
Therefore, the solution is:
[tex]\[ \boxed{80} \][/tex]
[tex]\[ \sqrt[3]{9 + \sqrt{x}} + \sqrt[3]{9 - \sqrt{x}} = 3 \][/tex]
To find [tex]\( x \)[/tex], let's start by making a substitution:
[tex]\[ a = \sqrt[3]{9 + \sqrt{x}} \quad \text{and} \quad b = \sqrt[3]{9 - \sqrt{x}} \][/tex]
Thus, the equation becomes:
[tex]\[ a + b = 3 \][/tex]
Next, let's cube both sides of the equation to eliminate the cube roots. Recall the binomial expansion:
[tex]\[ (a + b)^3 = a^3 + b^3 + 3ab(a + b) \][/tex]
Since [tex]\( a + b = 3 \)[/tex], we can substitute that into the expanded form:
[tex]\[ 3^3 = a^3 + b^3 + 3ab \cdot 3 \][/tex]
Simplify:
[tex]\[ 27 = a^3 + b^3 + 9ab \][/tex]
We also know:
[tex]\[ a^3 = 9 + \sqrt{x} \quad \text{and} \quad b^3 = 9 - \sqrt{x} \][/tex]
So,
[tex]\[ a^3 + b^3 = (9 + \sqrt{x}) + (9 - \sqrt{x}) = 18 \][/tex]
Substitute [tex]\( a^3 + b^3 = 18 \)[/tex] into the simplified cubic equation:
[tex]\[ 27 = 18 + 9ab \][/tex]
Subtract 18 from both sides:
[tex]\[ 9 = 9ab \][/tex]
Divide both sides by 9:
[tex]\[ 1 = ab \][/tex]
Now, use the fact that [tex]\( ab = 1 \)[/tex] to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]. Recall the relationships:
[tex]\[ ab = 1 \][/tex]
[tex]\[ a + b = 3 \][/tex]
We have a system of equations:
1. [tex]\( a + b = 3 \)[/tex]
2. [tex]\( ab = 1 \)[/tex]
These are the roots of the quadratic equation:
[tex]\[ t^2 - (a + b)t + ab = 0 \][/tex]
Replace [tex]\( a + b \)[/tex] and [tex]\( ab \)[/tex]:
[tex]\[ t^2 - 3t + 1 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ t = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus,
[tex]\[ a = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad b = \frac{3 - \sqrt{5}}{2} \][/tex]
Remember that:
[tex]\[ a^3 = 9 + \sqrt{x} \quad \text{and} \quad b^3 = 9 - \sqrt{x} \][/tex]
Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] back:
[tex]\[ a^3 = \left( \frac{3 + \sqrt{5}}{2} \right)^3 \][/tex]
[tex]\[ b^3 = \left( \frac{3 - \sqrt{5}}{2} \right)^3 \][/tex]
We now need to compute these cubes:
[tex]\[ \left( \frac{3 + \sqrt{5}}{2} \right)^3 = \frac{(3 + \sqrt{5})^3}{8} \][/tex]
[tex]\[ (3 + \sqrt{5})^3 = 27 + 3 \cdot 9 \cdot \sqrt{5} + 3 \cdot 5 + \sqrt{5}^3 = 27 + 27\sqrt{5} + 15 + 5\sqrt{5} \][/tex]
[tex]\[ = 42 + 32\sqrt{5} \][/tex]
So,
[tex]\[ \left( \frac{3 + \sqrt{5}}{2} \right)^3 = \frac{42 + 32\sqrt{5}}{8} = \frac{21 + 16\sqrt{5}}{4} \][/tex]
Similarly, for [tex]\( b \)[/tex]:
[tex]\[ \left( \frac{3 - \sqrt{5}}{2} \right)^3 = \frac{(3 - \sqrt{5})^3}{8} \][/tex]
[tex]\[ (3 - \sqrt{5})^3 = 27 - 27\sqrt{5} + 15 - 5\sqrt{5} = 42 - 32\sqrt{5} \][/tex]
So,
[tex]\[ \left( \frac{3 - \sqrt{5}}{2} \right)^3 = \frac{42 - 32\sqrt{5}}{8} = \frac{21 - 16\sqrt{5}}{4} \][/tex]
We have:
[tex]\[ 9 + \sqrt{x} = \frac{21 + 16\sqrt{5}}{4} \][/tex]
[tex]\[ 9 - \sqrt{x} = \frac{21 - 16\sqrt{5}}{4} \][/tex]
Subtract these equations:
[tex]\[ (9 + \sqrt{x}) - (9 - \sqrt{x}) = \frac{21 + 16\sqrt{5} - (21 - 16\sqrt{5})}{4} \][/tex]
[tex]\[ 2\sqrt{x} = \frac{32\sqrt{5}}{4} = 8\sqrt{5} \][/tex]
[tex]\[ \sqrt{x} = 4\sqrt{5} \][/tex]
Square both sides to find [tex]\( x \)[/tex]:
[tex]\[ x = (4\sqrt{5})^2 = 16 \times 5 = 80 \][/tex]
Therefore, the solution is:
[tex]\[ \boxed{80} \][/tex]
