Solving for $x + \frac{1}{x}$ using Algebraic Identity
We are given the equation $x^3 + \frac{1}{x^3} = 18$. We need to find the value of $x + \frac{1}{x}$.
We use the algebraic identity for the cube of a binomial:
$ \left(a + b\right)^3 = a^3 + b^3 + 3ab(a + b) $
Let $a = x$ and $b = \frac{1}{x}$. Substituting these into the identity:
$ \left(x + \frac{1}{x}\right)^3 = x^3 + \left(\frac{1}{x}\right)^3 + 3 \cdot x \cdot \frac{1}{x} \left(x + \frac{1}{x}\right) $
Simplify the equation:
$ \left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3 \left(x + \frac{1}{x}\right) $
Substituting the Given Value
We know that $x^3 + \frac{1}{x^3} = 18$. Substitute this value into the expanded equation:
$ \left(x + \frac{1}{x}\right)^3 = 18 + 3 \left(x + \frac{1}{x}\right) $
Solving the Cubic Equation
To simplify, let $y = x + \frac{1}{x}$. The equation becomes:
$ y^3 = 18 + 3y $
Rearrange the terms to form a cubic equation:
$ y^3 - 3y - 18 = 0 $
We can solve this cubic equation by testing integer factors of the constant term (-18). Let's test $y=3$:
$ (3)^3 - 3(3) - 18 = 27 - 9 - 18 = 18 - 18 = 0 $
Since $y=3$ satisfies the equation, it is a root.
Therefore, the value of $y$, which represents $x + \frac{1}{x}$, is 3.
Similar Questions
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Then the value of $\frac{3x}{2x^2 + 2 - 5x}$ will be: - If $3a + 4b = 2$ and $ab = \frac{1}{36}$, then $27a^3 + 64b^3$ is:
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Important Questions from Algebra
If 2x – y = 2 and xy = \(\frac{3}{2}\) , then what is the value of x 3– \(\frac{{{y^3}}}{8}\) ?
If (10a 3+ 4b 3) : (11a 3- 15b 3) = 7 : 5, then (3a + 5b) : (9a - 2b) =?
- If 4sin 2 θ = 3(1+ cos θ), 0° < θ < 90°, then what is the value of (2tan θ + 4sin θ - sec θ)?
The value of:
\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)
If (x + y) 3+ 27(x - y) 3= (Ax - 2y)(Bx 2+ Cxy + 13y 2), then the value of A - B - C is:
