If \(\left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47\) , what is the value of \(\left( {{x^6} + \frac{1}{{{x^6}}}} \right)?\)
18
Solving for \(x^6 + \frac{1}{x^6}\) from \(x^8 + \frac{1}{x^8}\)
This problem asks us to find the value of a specific algebraic expression involving powers of \(x\), given the value of another similar expression with different powers. We are given \( \left( {{x^8} + \frac{1}{{{x^8}}}} \right) = 47 \) and need to find \( \left( {{x^6} + \frac{1}{{{x^6}}}} \right) \).
To solve this, we can use standard algebraic identities related to terms of the form \( {{a^n} + \frac{1}{{{a^n}}}} \). A key identity is the square of a binomial:
\( \left( {a + \frac{1}{a}} \right)^2 = a^2 + 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2} \)
This can be rearranged to find a lower power:
\( a^2 + \frac{1}{a^2} = \left( {a + \frac{1}{a}} \right)^2 - 2 \)
Alternatively, to find \( a + \frac{1}{a} \) from \( a^2 + \frac{1}{a^2} \):
\( \left( {a + \frac{1}{a}} \right)^2 = a^2 + \frac{1}{a^2} + 2 \)
\( a + \frac{1}{a} = \sqrt{a^2 + \frac{1}{a^2} + 2} \) (Assuming \(a\) is real and \(a + \frac{1}{a}\) is positive).
Step-by-Step Calculation
We start with the highest power given:
\( x^8 + \frac{1}{x^8} = 47 \)
Let \( a = x^4 \). Then \( a^2 = x^8 \). We have \( a^2 + \frac{1}{a^2} = 47 \). We can find \( a + \frac{1}{a} \) using the identity:
\( \left( x^4 + \frac{1}{x^4} \right)^2 = x^8 + \frac{1}{x^8} + 2 \)
Substitute the given value:
\( \left( x^4 + \frac{1}{x^4} \right)^2 = 47 + 2 = 49 \)
Taking the square root (and assuming the positive value for simplicity in this context):
\( x^4 + \frac{1}{x^4} = \sqrt{49} = 7 \)
Now we have \( x^4 + \frac{1}{x^4} = 7 \). We can repeat the process to find \( x^2 + \frac{1}{x^2} \).
Let \( b = x^2 \). Then \( b^2 = x^4 \). We have \( b^2 + \frac{1}{b^2} = 7 \). We can find \( b + \frac{1}{b} \):
\( \left( x^2 + \frac{1}{x^2} \right)^2 = x^4 + \frac{1}{x^4} + 2 \)
Substitute the value we just found:
\( \left( x^2 + \frac{1}{x^2} \right)^2 = 7 + 2 = 9 \)
Taking the square root (and assuming the positive value):
\( x^2 + \frac{1}{x^2} = \sqrt{9} = 3 \)
We need to find \( x^6 + \frac{1}{x^6} \). We can relate this to \( x^2 + \frac{1}{x^2} \).
Let \( c = x^2 \). We need to find \( c^3 + \frac{1}{c^3} \). We know \( c + \frac{1}{c} = x^2 + \frac{1}{x^2} = 3 \).
We use the identity for the sum of cubes of a binomial:
\( \left( {a + b} \right)^3 = a^3 + b^3 + 3ab(a+b) \)
Rearranging for \( a^3 + b^3 \):
\( a^3 + b^3 = \left( {a + b} \right)^3 - 3ab(a+b) \)
Let \( a = x^2 \) and \( b = \frac{1}{x^2} \). Then \( ab = x^2 \cdot \frac{1}{x^2} = 1 \).
So, \( x^6 + \frac{1}{x^6} = (x^2)^3 + \left(\frac{1}{x^2}\right)^3 \)
\( x^6 + \frac{1}{x^6} = \left( x^2 + \frac{1}{x^2} \right)^3 - 3 \left( x^2 \cdot \frac{1}{x^2} \right) \left( x^2 + \frac{1}{x^2} \right) \)
\( x^6 + \frac{1}{x^6} = \left( x^2 + \frac{1}{x^2} \right)^3 - 3 \left( x^2 + \frac{1}{x^2} \right) \)
We found that \( x^2 + \frac{1}{x^2} = 3 \). Substitute this value:
\( x^6 + \frac{1}{x^6} = (3)^3 - 3(3) \)
\( x^6 + \frac{1}{x^6} = 27 - 9 \)
\( x^6 + \frac{1}{x^6} = 18 \)
Thus, the value of \( \left( {{x^6} + \frac{1}{{{x^6}}}} \right) \) is 18.
Algebraic Identities Revision Table
| Expression | Related Identity |
|---|---|
| \( a^2 + \frac{1}{a^2} \) | \( \left( a + \frac{1}{a} \right)^2 - 2 \) |
| \( a + \frac{1}{a} \) | \( \sqrt{a^2 + \frac{1}{a^2} + 2} \) |
| \( a^3 + \frac{1}{a^3} \) | \( \left( a + \frac{1}{a} \right)^3 - 3 \left( a + \frac{1}{a} \right) \) |
Additional Information on Algebraic Problem Solving
Problems involving powers of \( x + \frac{1}{x} \) are common in algebra and competitive exams. Recognizing the pattern and the related identities is crucial for solving them efficiently. The core idea is often to step down or step up the powers using squaring or cubing identities.
In this problem, we went from \(x^8\) to \(x^4\), then to \(x^2\). Then, we used the value of the \(x^2\) expression to find the value of the \(x^6\) expression. This step-wise approach is typical.
We assumed positive roots when taking the square root (e.g., from \((x^4 + 1/x^4)^2 = 49\) to \(x^4 + 1/x^4 = 7\)). If \(x\) were complex or negative such that \(x^n + 1/x^n\) could be negative, we might consider negative roots as well, leading to multiple possible values. However, in standard problems like this, especially with positive values given for higher even powers, it's usually implied that we consider real \(x\) where the positive root applies for intermediate steps.
Similar Questions
What is the HCF of acx3 + bcx2 + adx2 + acdx + bdx + bcd and adx3 + acx2 + bdx2 + bcx + acdx + bcd if HCF (c, d) = 1, c ≠ d?
If 2s = a + b + c, then what is s2 + (s - a)(s - b) + (s - b)(s - c) + (s - c)(s - a) equal to ?
If 2x - 3y - 7 = 0, then what is the value of 8x 3 - 36x 2y + 54xy 2 - 27y 3 - 340 ?
If \(A + B = \rm \frac{x^2 - 8}{x + 2} \ \ and \ A - B = \frac{-x^2 + 2x + 4}{x + 2}\) then what is B equal to ?
If \(96 - 64a^3 + \frac{8}{a^6} - \frac{48}{a^3 } - t^3 = 0\) then what is a 2t + 4a 3 equal to ?
The sum of all possible products taken two at a time out of the numbers \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 5\) is
The sum of all possible products taken two at a time out of the numbers ± 1, ± 2, ±3, ± 4 is
If \(\rm\frac{{61}}{{19}}{\rm{}} = {\rm{}}3{\rm{\;}} + {\rm{\;}}\frac{1}{{x\; + \;\frac{1}{{y\; + \;\frac{1}{z}}}}}\) where x, y and z are natural numbers, then what is z equal to?
- If (s − a) + (s − b) + (s − c) = s, then the value of \(\rm\frac{(s−a)^2+(s−b)^2+(s−c)^2+s^2}{a^2+b^2+c^2}\) will be
If x = 2 1/3 + 2 -1/3 , then the value of 2x 3- 6x - 5 is equal to
Important Questions from Identities
Simplify.
\(\frac{2.5 \times 2.5 \times 2.5-1.5 \times 1.5 \times 1.5}{2.5 \times 2.5+2.5 \times 1.5+1.5 \times 1.5}\)
If \(2x + { {1} \over 3x}= 5, x ≠ 0\) , then what is the value of \(27x^3+{{1} \over 8x^3}\) ?
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:
- \(\dfrac{(0.73)^3+(0.31)^3}{(0.73)^2-0.73\times0.31+(0.31)^2}\)
- \(\dfrac{(5.17-2.19)^2-(5.17+2.19)^2}{11.3223}\)
