If \((x + {1\over x}) = \sqrt6\), and x > 1, what is the value of \((x^8 - {1 \over x^8})\)?
112√3
Understanding the Algebraic Expression
The problem asks us to find the value of a complex algebraic expression, \((x^8 - {1 \over x^8})\), given a simpler relationship involving \(x\), specifically \((x + {1\over x}) = \sqrt6\) and that \(x > 1\). To solve this, we can use algebraic identities to break down the complex expression into simpler parts that can be calculated from the given information.
Strategy for Solving the \(x^8\) Problem
The expression \((x^8 - {1 \over x^8})\) is a difference of squares. We can repeatedly apply the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\), to simplify it:
\[x^8 - {1 \over x^8} = \left(x^4\right)^2 - \left({1 \over x^4}\right)^2 = \left(x^4 - {1 \over x^4}\right)\left(x^4 + {1 \over x^4}\right)\]
We can continue factoring the difference term:
\[x^4 - {1 \over x^4} = \left(x^2\right)^2 - \left({1 \over x^2}\right)^2 = \left(x^2 - {1 \over x^2}\right)\left(x^2 + {1 \over x^2}\right)\]
And again:
\[x^2 - {1 \over x^2} = (x)^2 - \left({1 \over x}\right)^2 = \left(x - {1 \over x}\right)\left(x + {1 \over x}\right)\]
Substituting these back, we get:
\[x^8 - {1 \over x^8} = \left(x - {1 \over x}\right)\left(x + {1 \over x}\right)\left(x^2 + {1 \over x^2}\right)\left(x^4 + {1 \over x^4}\right)\]
Now, we need to find the values of each of the terms on the right side using the given \((x + {1\over x}) = \sqrt6\).
Calculating Required Terms from \(x + {1 \over x}\)
Finding \(x^2 + {1 \over x^2}\)
We can use the identity \((a+b)^2 = a^2 + 2ab + b^2\). Let \(a=x\) and \(b={1 \over x}\):
\[\left(x + {1 \over x}\right)^2 = x^2 + 2 \cdot x \cdot {1 \over x} + \left({1 \over x}\right)^2\]
\[\left(x + {1 \over x}\right)^2 = x^2 + 2 + {1 \over x^2}\]
Rearranging to find \(x^2 + {1 \over x^2}\):
\[x^2 + {1 \over x^2} = \left(x + {1 \over x}\right)^2 - 2\]
Substitute the given value \((x + {1\over x}) = \sqrt6\):
\[x^2 + {1 \over x^2} = (\sqrt6)^2 - 2\]
\[x^2 + {1 \over x^2} = 6 - 2\]
\[x^2 + {1 \over x^2} = 4\]
Finding \(x^4 + {1 \over x^4}\)
We can use the same identity again, but with \(a=x^2\) and \(b={1 \over x^2}\):
\[\left(x^2 + {1 \over x^2}\right)^2 = (x^2)^2 + 2 \cdot x^2 \cdot {1 \over x^2} + \left({1 \over x^2}\right)^2\]
\[\left(x^2 + {1 \over x^2}\right)^2 = x^4 + 2 + {1 \over x^4}\]
Rearranging to find \(x^4 + {1 \over x^4}\):
\[x^4 + {1 \over x^4} = \left(x^2 + {1 \over x^2}\right)^2 - 2\]
Substitute the value of \(x^2 + {1 \over x^2}\) we just found (which is 4):
\[x^4 + {1 \over x^4} = (4)^2 - 2\]
\[x^4 + {1 \over x^4} = 16 - 2\]
\[x^4 + {1 \over x^4} = 14\]
Finding \(x - {1 \over x}\) using the Condition \(x > 1\)
We can use the identity \((a-b)^2 = a^2 - 2ab + b^2\). Let \(a=x\) and \(b={1 \over x}\):
\[\left(x - {1 \over x}\right)^2 = x^2 - 2 \cdot x \cdot {1 \over x} + \left({1 \over x}\right)^2\]
\[\left(x - {1 \over x}\right)^2 = x^2 - 2 + {1 \over x^2}\]
Notice that \(x^2 + {1 \over x^2}\) is a term we have already calculated (it is 4). So:
\[\left(x - {1 \over x}\right)^2 = \left(x^2 + {1 \over x^2}\right) - 2\]
Substitute the value of \(x^2 + {1 \over x^2}\):
\[\left(x - {1 \over x}\right)^2 = 4 - 2\]
\[\left(x - {1 \over x}\right)^2 = 2\]
Taking the square root of both sides:
\[x - {1 \over x} = \pm \sqrt{2}\]
The problem states that \(x > 1\). If \(x > 1\), then \(0 < {1 \over x} < 1\). This means that \(x\) is larger than \({1 \over x}\), so their difference \(x - {1 \over x}\) must be positive.
Therefore, we take the positive square root:
\[x - {1 \over x} = \sqrt{2}\]
Calculating the Final Value of \(x^8 - {1 \over x^8}\)
Now we have all the parts needed for the factored expression:
- \(\left(x - {1 \over x}\right) = \sqrt{2}\)
- \(\left(x + {1 \over x}\right) = \sqrt6\) (Given)
- \(\left(x^2 + {1 \over x^2}\right) = 4\)
- \(\left(x^4 + {1 \over x^4}\right) = 14\)
Substitute these values into the factored expression for \(x^8 - {1 \over x^8}\):
\[x^8 - {1 \over x^8} = \left(x - {1 \over x}\right)\left(x + {1 \over x}\right)\left(x^2 + {1 \over x^2}\right)\left(x^4 + {1 \over x^4}\right)\]
\[x^8 - {1 \over x^8} = (\sqrt{2}) \cdot (\sqrt{6}) \cdot (4) \cdot (14)\]
Multiply the terms:
\[x^8 - {1 \over x^8} = (\sqrt{2 \cdot 6}) \cdot (4 \cdot 14)\]
\[x^8 - {1 \over x^8} = (\sqrt{12}) \cdot (56)\]
Simplify the square root \(\sqrt{12}\): \(\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}\).
\[x^8 - {1 \over x^8} = (2\sqrt{3}) \cdot (56)\]
\[x^8 - {1 \over x^8} = 112\sqrt{3}\]
Final Answer Calculation
Based on the steps, the value of \((x^8 - {1 \over x^8})\) is \(112\sqrt{3}\).
Revision Table: Key Algebraic Identities
Here's a quick summary of the identities used in this problem:
| Identity | Formula |
|---|---|
| Square of a sum | \((a+b)^2 = a^2 + 2ab + b^2\) |
| Square of a difference | \((a-b)^2 = a^2 - 2ab + b^2\) |
| Difference of squares | \(a^2 - b^2 = (a-b)(a+b)\) |
Additional Information on Powers and Reciprocals
Problems involving expressions like \(x^n + {1 \over x^n}\) or \(x^n - {1 \over x^n}\) are common in algebra. If you know the value of \(x + {1 \over x}\) or \(x - {1 \over x}\), you can iteratively find the values for higher powers using the identities discussed. For example, from \(x^2 + {1 \over x^2}\), you can find \(x^4 + {1 \over x^4}\), then \(x^8 + {1 \over x^8}\), and so on. Similarly, knowing \(x + {1 \over x}\) allows finding \(x - {1 \over x}\) (and vice versa) using the relationship \((x + {1 \over x})^2 - (x - {1 \over x})^2 = 4\). Always pay attention to conditions like \(x > 1\) or \(0 < x < 1\) as they help determine the sign when taking square roots.
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