\({\rm{If}}\;{a^2} + \frac{1}{{{a^2}}} = 3,\;{\rm{then}}\;{a^3} + \frac{1}{{{a^3}}} = ?\)
2√5
Understanding the Algebraic Problem
The question asks us to find the value of the expression \(a^3 + \frac{1}{a^3}\) given the value of \(a^2 + \frac{1}{a^2}\).
This type of problem involves algebraic identities relating powers of a variable and its reciprocal. We can use standard formulas for squares and cubes of sums to solve this.
Step-by-Step Solution for Finding \(a^3 + \frac{1}{a^3}\)
We are given:
\({a^2} + \frac{1}{{{a^2}}} = 3\)
We need to find \(a^3 + \frac{1}{{{a^3}}}\).
Step 1: Finding the value of \(a + \frac{1}{a}\)
We know the identity:
\[{\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} + 2\]Using this identity with \(x=a\), we get:
\[{\left( {a + \frac{1}{a}} \right)^2} = {a^2} + \frac{1}{{{a^2}}} + 2\]Substitute the given value \(a^2 + \frac{1}{a^2} = 3\):
\[{\left( {a + \frac{1}{a}} \right)^2} = 3 + 2\] \[{\left( {a + \frac{1}{a}} \right)^2} = 5\]Taking the square root of both sides, we get:
\[a + \frac{1}{a} = \pm \sqrt 5\]For this problem, the options are positive values, which implies we should consider the positive root for \(a + \frac{1}{a}\). Thus, we take \(a + \frac{1}{a} = \sqrt 5\).
Step 2: Finding the value of \(a^3 + \frac{1}{a^3}\)
We know another identity related to the cube of a sum:
\[{\left( {x + \frac{1}{x}} \right)^3} = {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right)\]Rearranging this identity to find \(x^3 + \frac{1}{x^3}\):
\[{x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right)\]Using this identity with \(x=a\), we get:
\[{a^3} + \frac{1}{{{a^3}}} = {\left( {a + \frac{1}{a}} \right)^3} - 3\left( {a + \frac{1}{a}} \right)\]Now, substitute the value \(a + \frac{1}{a} = \sqrt 5\) that we found in Step 1:
\[{a^3} + \frac{1}{{{a^3}}} = {\left( {\sqrt 5 } \right)^3} - 3\left( {\sqrt 5 } \right)\]Calculate \({\left( {\sqrt 5 } \right)^3}\):
\[{\left( {\sqrt 5 } \right)^3} = \sqrt 5 \times \sqrt 5 \times \sqrt 5 = 5\sqrt 5\]Substitute this back into the equation for \(a^3 + \frac{1}{a^3}\):
\[{a^3} + \frac{1}{{{a^3}}} = 5\sqrt 5 - 3\sqrt 5\]Perform the subtraction:
\[{a^3} + \frac{1}{{{a^3}}} = \left( {5 - 3} \right)\sqrt 5\] \[{a^3} + \frac{1}{{{a^3}}} = 2\sqrt 5\]The value of \(a^3 + \frac{1}{a^3}\) is \(2\sqrt 5\).
Analyzing the Options
Let's compare our result with the given options:
- Option 1: \(2\sqrt 3\)
- Option 2: \(2\sqrt 5\)
- Option 3: \(3\sqrt 3\)
- Option 4: \(3\sqrt 5\)
Our calculated value \(2\sqrt 5\) matches Option 2.
| Given | To Find | Key Identity 1 | Key Identity 2 | Calculated Value |
|---|---|---|---|---|
| \(a^2 + \frac{1}{a^2} = 3\) | \(a^3 + \frac{1}{a^3}\) | \({\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} + 2\) | \({\left( {x + \frac{1}{x}} \right)^3} = {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right)\) | \(2\sqrt 5\) |
Revision Table: Understanding Related Algebraic Identities
| Identity | Formula | Usage |
|---|---|---|
| Square of a sum | \({\left( {x+y} \right)^2} = {x^2} + 2xy + {y^2}\) | General expansion |
| Square of a difference | \({\left( {x-y} \right)^2} = {x^2} - 2xy + {y^2}\) | General expansion |
| Square of sum (reciprocal) | \({\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} + 2\) | Relates sum and square of sum for reciprocals |
| Square of difference (reciprocal) | \({\left( {x - \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} - 2\) | Relates difference and square of difference for reciprocals |
| Cube of a sum | \({\left( {x+y} \right)^3} = {x^3} + 3{x^2}y + 3x{y^2} + {y^3} = {x^3} + {y^3} + 3xy(x+y)\) | General expansion |
| Cube of a difference | \({\left( {x-y} \right)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3} = {x^3} - {y^3} - 3xy(x-y)\) | General expansion |
| Cube of sum (reciprocal) | \({\left( {x + \frac{1}{x}} \right)^3} = {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right)\) | Relates sum and cube of sum for reciprocals |
| Cube of difference (reciprocal) | \({\left( {x - \frac{1}{x}} \right)^3} = {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right)\) | Relates difference and cube of difference for reciprocals |
Additional Information on Solving Algebraic Expressions
Problems involving powers like \(a^n + \frac{1}{a^n}\) or \(a^n - \frac{1}{a^n}\) can often be solved by finding the value of \(a + \frac{1}{a}\) or \(a - \frac{1}{a}\) first.
- If you need to find \(a^3 + \frac{1}{a^3}\), it's easiest if you know \(a + \frac{1}{a}\).
- If you need to find \(a^3 - \frac{1}{a^3}\), it's easiest if you know \(a - \frac{1}{a}\).
- You can switch between finding \(a + \frac{1}{a}\) and \(a - \frac{1}{a}\) if you know \(a^2 + \frac{1}{a^2}\), because \({\left( {a + \frac{1}{a}} \right)^2} = a^2 + \frac{1}{a^2} + 2\) and \({\left( {a - \frac{1}{a}} \right)^2} = a^2 + \frac{1}{a^2} - 2\).
- Higher powers can be found by repeatedly applying these identities or by finding a pattern. For example, \(a^4 + \frac{1}{a^4}\) can be found from \(a^2 + \frac{1}{a^2}\) using the squaring identity: \({\left( {{a^2} + \frac{1}{{{a^2}}}} \right)^2} = {a^4} + \frac{1}{{{a^4}}} + 2\).
These techniques are fundamental in algebraic manipulation and useful in various mathematical problems.
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