Finding the Value of a³ + 1/a³

We are given the equation:

$\qquad a + \frac{1}{a} = 1$

We need to find the value of the expression:

$\qquad a^3 + \frac{1}{a^3}$

Using the Algebraic Identity for Cubes

To find the value of $a^3 + \frac{1}{a^3}$, we can use a standard algebraic identity related to the sum of cubes. The identity we will use is the expansion of $(x+y)^3$:

$\qquad (x+y)^3 = x^3 + y^3 + 3xy(x+y)$

In our case, we can substitute $x=a$ and $y=\frac{1}{a}$ into this identity. Let's apply this to the given equation $a + \frac{1}{a} = 1$ by cubing both sides.

Calculation Steps to Find a³ + 1/a³

Let's cube both sides of the given equation $a + \frac{1}{a} = 1$:

$\qquad \left(a + \frac{1}{a}\right)^3 = (1)^3$

Using the identity $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$ with $x=a$ and $y=\frac{1}{a}$ on the left side:

$\qquad a^3 + \left(\frac{1}{a}\right)^3 + 3 \cdot a \cdot \frac{1}{a} \left(a + \frac{1}{a}\right) = 1^3$

Simplify the terms:

• $(\frac{1}{a})^3 = \frac{1^3}{a^3} = \frac{1}{a^3}$
• $a \cdot \frac{1}{a} = \frac{a}{a} = 1$ (assuming $a \neq 0$, which must be true for $\frac{1}{a}$ to exist and the original equation to hold)
• $1^3 = 1$

Substituting these simplified terms back into the equation:

$\qquad a^3 + \frac{1}{a^3} + 3 \cdot 1 \cdot \left(a + \frac{1}{a}\right) = 1$

This simplifies to:

$\qquad a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) = 1$

Now, we know from the original given condition that $a + \frac{1}{a} = 1$. Substitute this value into the equation:

$\qquad a^3 + \frac{1}{a^3} + 3(1) = 1$

$\qquad a^3 + \frac{1}{a^3} + 3 = 1$

To find the value of $a^3 + \frac{1}{a^3}$, subtract 3 from both sides of the equation:

$\qquad a^3 + \frac{1}{a^3} = 1 - 3$

$\qquad a^3 + \frac{1}{a^3} = -2$

Conclusion on the Value of a³ + 1/a³

The value of $a^3 + \frac{1}{a^3}$ is -2.

Revision Table: Common Algebraic Identities

Additional Information: Understanding a + 1/a = 1

The equation $a + \frac{1}{a} = 1$ can be rearranged by multiplying by $a$ (assuming $a \neq 0$):

$\qquad a^2 + 1 = a$

Rearranging into a quadratic equation:

$\qquad a^2 - a + 1 = 0$

The roots of this quadratic equation can be found using the quadratic formula $a = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Here, $a=1$, $b=-1$, $c=1$.

The discriminant is $\Delta = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3$.

Since the discriminant is negative, the roots are complex numbers:

$\qquad a = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}$

These are related to the complex cube roots of -1. Specifically, if $\omega = e^{i\pi/3} = \frac{1+i\sqrt{3}}{2}$, then $\omega^2 = e^{i2\pi/3} = \frac{-1+i\sqrt{3}}{2}$, $\omega^3 = -1$, $\omega^6 = 1$. The roots of $a^2 - a + 1 = 0$ are related to these complex roots. For example, one root is $\frac{1+i\sqrt{3}}{2}$. Cubing this complex number would indeed yield -1. The other root is $\frac{1-i\sqrt{3}}{2}$, which is the complex conjugate. Cubing this also yields -1. Since $a^3 = -1$, then $1/a^3 = 1/(-1) = -1$. Therefore, $a^3 + 1/a^3 = -1 + (-1) = -2$. This confirms the result obtained using the algebraic identity without needing to calculate the specific value of 'a'.