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To find the value of [tex]\( a^3 + \frac{1}{a^3} \)[/tex] given that [tex]\( a^2 + \frac{1}{a^2} = 18 \)[/tex] and using only the positive value of [tex]\( a - \frac{1}{a} \)[/tex], we can follow these steps:
1. Find [tex]\( a \)[/tex] from [tex]\( a^2 + \frac{1}{a^2} = 18 \)[/tex]:
- We start with the equation [tex]\( a^2 + \frac{1}{a^2} = 18 \)[/tex].
- Denote [tex]\( x = a^2 \)[/tex]. Thus, the equation becomes [tex]\( x + \frac{1}{x} = 18 \)[/tex].
2. Solve for [tex]\( x \)[/tex]:
- Multiply both sides by [tex]\( x \)[/tex] to clear the fraction:
[tex]\[ x^2 + 1 = 18x \implies x^2 - 18x + 1 = 0 \][/tex]
- Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -18 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ x = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{18 \pm \sqrt{324 - 4}}{2} = \frac{18 \pm \sqrt{320}}{2} = \frac{18 \pm 4\sqrt{20}}{2} = 9 \pm 2\sqrt{5} \][/tex]
- Since [tex]\( x = a^2 \geq 0 \)[/tex], and [tex]\( 9 - 2\sqrt{5} \)[/tex] is a positive number, we have [tex]\( x = 9 + 2\sqrt{5} \)[/tex].
3. Find the positive value of [tex]\( a - \frac{1}{a} \)[/tex]:
- We know that [tex]\( a = \sqrt{x} \)[/tex], so [tex]\( a = \sqrt{9 + 2\sqrt{5}} \)[/tex].
- To work with [tex]\( \sqrt{x} \)[/tex], we rationalize [tex]\( \sqrt{x} = 2 + \sqrt{5} \)[/tex]. Thus, [tex]\( a = 2 + \sqrt{5} \)[/tex].
- Then calculate [tex]\( \frac{1}{a} \)[/tex]:
[tex]\[ \frac{1}{a} = \frac{1}{2 + \sqrt{5}} = 2 - \sqrt{5} \][/tex]
- Therefore,
[tex]\[ a - \frac{1}{a} = (2 + \sqrt{5}) - (2 - \sqrt{5}) = 2\sqrt{5} \][/tex]
4. Calculate [tex]\( a^3 + \frac{1}{a^3} \)[/tex]:
- We use the identity:
[tex]\[ \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \][/tex]
- Given that:
[tex]\[ a^2 + \frac{1}{a^2} = 18 \quad \text{implies} \quad \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 = 18 + 2 = 20 \][/tex]
- Thus, [tex]\( a + \frac{1}{a} = \sqrt{20} = 2\sqrt{5} \)[/tex]:
[tex]\[ \left(2 + \sqrt{5} + 2 - \sqrt{5}\right)^3 = \left(2\sqrt{5}\right)^3 = 34\sqrt{5} \][/tex]
- Hence,
[tex]\[ a^3 + \frac{1}{a^3} = 34\sqrt{5} \][/tex]
Thus, the value of [tex]\( a^3 + \frac{1}{a^3} \)[/tex] is [tex]\( 34\sqrt{5} \)[/tex].
1. Find [tex]\( a \)[/tex] from [tex]\( a^2 + \frac{1}{a^2} = 18 \)[/tex]:
- We start with the equation [tex]\( a^2 + \frac{1}{a^2} = 18 \)[/tex].
- Denote [tex]\( x = a^2 \)[/tex]. Thus, the equation becomes [tex]\( x + \frac{1}{x} = 18 \)[/tex].
2. Solve for [tex]\( x \)[/tex]:
- Multiply both sides by [tex]\( x \)[/tex] to clear the fraction:
[tex]\[ x^2 + 1 = 18x \implies x^2 - 18x + 1 = 0 \][/tex]
- Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -18 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ x = \frac{18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{18 \pm \sqrt{324 - 4}}{2} = \frac{18 \pm \sqrt{320}}{2} = \frac{18 \pm 4\sqrt{20}}{2} = 9 \pm 2\sqrt{5} \][/tex]
- Since [tex]\( x = a^2 \geq 0 \)[/tex], and [tex]\( 9 - 2\sqrt{5} \)[/tex] is a positive number, we have [tex]\( x = 9 + 2\sqrt{5} \)[/tex].
3. Find the positive value of [tex]\( a - \frac{1}{a} \)[/tex]:
- We know that [tex]\( a = \sqrt{x} \)[/tex], so [tex]\( a = \sqrt{9 + 2\sqrt{5}} \)[/tex].
- To work with [tex]\( \sqrt{x} \)[/tex], we rationalize [tex]\( \sqrt{x} = 2 + \sqrt{5} \)[/tex]. Thus, [tex]\( a = 2 + \sqrt{5} \)[/tex].
- Then calculate [tex]\( \frac{1}{a} \)[/tex]:
[tex]\[ \frac{1}{a} = \frac{1}{2 + \sqrt{5}} = 2 - \sqrt{5} \][/tex]
- Therefore,
[tex]\[ a - \frac{1}{a} = (2 + \sqrt{5}) - (2 - \sqrt{5}) = 2\sqrt{5} \][/tex]
4. Calculate [tex]\( a^3 + \frac{1}{a^3} \)[/tex]:
- We use the identity:
[tex]\[ \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \][/tex]
- Given that:
[tex]\[ a^2 + \frac{1}{a^2} = 18 \quad \text{implies} \quad \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 = 18 + 2 = 20 \][/tex]
- Thus, [tex]\( a + \frac{1}{a} = \sqrt{20} = 2\sqrt{5} \)[/tex]:
[tex]\[ \left(2 + \sqrt{5} + 2 - \sqrt{5}\right)^3 = \left(2\sqrt{5}\right)^3 = 34\sqrt{5} \][/tex]
- Hence,
[tex]\[ a^3 + \frac{1}{a^3} = 34\sqrt{5} \][/tex]
Thus, the value of [tex]\( a^3 + \frac{1}{a^3} \)[/tex] is [tex]\( 34\sqrt{5} \)[/tex].
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