If x = 2 1/3 + 2 -1/3 , then the value of 2x 3- 6x - 5 is equal to
0
Evaluating Algebraic Expressions with Exponents
The question asks us to find the value of the expression \( 2x^3 - 6x - 5 \), given that \( x = 2^{1/3} + 2^{-1/3} \).
Let's break down the problem step-by-step.
Step 1: Understand the Given Value of x
We are given the value of \( x \) as a sum of two terms involving fractional exponents:
\( x = 2^{1/3} + 2^{-1/3} \)
This expression looks like it could simplify nicely when cubed.
Step 2: Cube the Expression for x
To evaluate \( 2x^3 - 6x - 5 \), we first need to find \( x^3 \). We can cube both sides of the equation for \( x \).
\( x^3 = (2^{1/3} + 2^{-1/3})^3 \)
We use the algebraic identity for cubing a sum: \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \).
Let \( a = 2^{1/3} \) and \( b = 2^{-1/3} \).
Then,
- \( a^3 = (2^{1/3})^3 = 2^{(1/3) \times 3} = 2^1 = 2 \)
- \( b^3 = (2^{-1/3})^3 = 2^{(-1/3) \times 3} = 2^{-1} = \frac{1}{2} \)
- \( ab = (2^{1/3})(2^{-1/3}) = 2^{1/3 + (-1/3)} = 2^{1/3 - 1/3} = 2^0 = 1 \)
- \( a+b = 2^{1/3} + 2^{-1/3} = x \) (from the original definition of x)
Substitute these values back into the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \):
\( x^3 = 2 + \frac{1}{2} + 3(1)(x) \)
\( x^3 = 2 + 0.5 + 3x \)
\( x^3 = 2.5 + 3x \)
We can also write 2.5 as \( \frac{5}{2} \):
\( x^3 = \frac{5}{2} + 3x \)
Step 3: Substitute x³ into the Target Expression
Now we substitute the expression for \( x^3 \) we found into the expression \( 2x^3 - 6x - 5 \).
The target expression is \( 2x^3 - 6x - 5 \).
Substitute \( x^3 = \frac{5}{2} + 3x \):
\( 2\left(\frac{5}{2} + 3x\right) - 6x - 5 \)
Step 4: Simplify the Expression
Now, distribute the 2 and combine like terms:
\( 2 \times \frac{5}{2} + 2 \times 3x - 6x - 5 \)
\( 5 + 6x - 6x - 5 \)
Combine the terms with \( x \):
\( (6x - 6x) + (5 - 5) \)
\( 0 + 0 \)
\( 0 \)
The value of the expression \( 2x^3 - 6x - 5 \) is 0.
Summary of Steps
Here is a quick overview of the process:
- Identify the given value of \( x \).
- Cube \( x \) using the formula \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \), substituting \( a=2^{1/3} \) and \( b=2^{-1/3} \).
- Simplify the expression for \( x^3 \).
- Substitute the simplified \( x^3 \) into the expression \( 2x^3 - 6x - 5 \).
- Simplify the resulting expression to find the final value.
Let's put the key intermediate result in a table.
| Expression | Result |
|---|---|
| \(x\) | \(2^{1/3} + 2^{-1/3}\) |
| \(a\) | \(2^{1/3}\) |
| \(b\) | \(2^{-1/3}\) |
| \(a^3\) | \(2\) |
| \(b^3\) | \(1/2\) |
| \(ab\) | \(1\) |
| \(x^3\) | \(2.5 + 3x\) or \(5/2 + 3x\) |
The final calculation yielded 0.
Revision Table - Key Concepts
Understanding the properties of exponents and algebraic identities is crucial for solving this type of problem.
| Concept | Description | Formula/Example |
|---|---|---|
| Fractional Exponents | \(a^{1/n}\) is the \(n\)-th root of \(a\). | \(2^{1/3}\) is the cube root of 2. |
| Negative Exponents | \(a^{-n} = 1/a^n\). | \(2^{-1/3} = 1/2^{1/3}\). |
| Product of Powers | \(a^m \cdot a^n = a^{m+n}\). | \(2^{1/3} \cdot 2^{-1/3} = 2^{1/3 - 1/3} = 2^0\). |
| Zero Exponent | \(a^0 = 1\) (for \(a \neq 0\)). | \(2^0 = 1\). |
| Cube of a Sum | \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \). | Used to expand \( (2^{1/3} + 2^{-1/3})^3 \). |
Additional Information - Related Problems
Problems involving expressions like \( x = a^{1/3} + a^{-1/3} \) often simplify nicely when finding expressions involving \( x^3 \). The key is the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \), where \(ab\) often simplifies to a constant when \( a \) and \( b \) are reciprocals with roots, like \( a^{1/n} \) and \( a^{-1/n} \).
Consider a general case: If \( x = a^{1/3} + a^{-1/3} \), then
\( x^3 = (a^{1/3} + a^{-1/3})^3 = (a^{1/3})^3 + (a^{-1/3})^3 + 3(a^{1/3})(a^{-1/3})(a^{1/3} + a^{-1/3}) \)
\( x^3 = a + a^{-1} + 3(a^0)(x) \)
\( x^3 = a + \frac{1}{a} + 3x \)
\( x^3 - 3x = a + \frac{1}{a} \)
In our specific problem, \( a=2 \). So \( x = 2^{1/3} + 2^{-1/3} \).
Then \( x^3 - 3x = 2 + \frac{1}{2} = \frac{5}{2} \).
From \( x^3 = \frac{5}{2} + 3x \), we have \( x^3 - 3x - \frac{5}{2} = 0 \).
The expression we needed to evaluate was \( 2x^3 - 6x - 5 \). We can factor out a 2:
\( 2(x^3 - 3x - \frac{5}{2}) \)
Since \( x^3 - 3x - \frac{5}{2} = 0 \), the expression becomes \( 2(0) = 0 \).
This confirms our earlier calculation and shows a common pattern for such problems.
Similar Questions
What is the square root of 4x 4 + 8x 3 - 4x + 1?
If x + 1/x = 8, then what is x 2 + 1/x 2 equal to?
If a + b = 5 and ab = 6, then what is the value of a 3+ b 3?
If \(x=a+b+\frac{(a-b)^2}{4 a+4 b}\) and \(y=\frac{a+b}{4}+\frac{a b}{a+b}\) then what is the value of (x - a)2 - (y - b)2 ?
The quotient of 8x 3– y 3when divided by 2xy + 4x 2+ y 2is
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?
Important Questions from Identities
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:
- \(\dfrac{(5.17-2.19)^2-(5.17+2.19)^2}{11.3223}\)
If a(a + b + c) 2= 1792; b(a + b + c) 2= 1536; c(a + b + c) 2= 768 then what will be the value of a?
If a + b + c = 5, a 2+ b 2+ c 2= 27, a 3 + b 3 + c 3 = 125, then the value of \(\frac{abc}{5}\) is:
(999) 2- 2 2=
