Question

Simplify 4/(1 - 3^(1/3) + 9^(1/3))

Answer 1

To simplify the expression \( \frac{4}{1 - 3^{1/3} + 9^{1/3}} \), we will start by rewriting \( 9^{1/3} \):

\[
9^{1/3} = (3^2)^{1/3} = 3^{2/3}
\]

This allows us to rewrite the expression in the denominator:

\[
1 - 3^{1/3} + 9^{1/3} = 1 - 3^{1/3} + 3^{2/3}
\]

Now let's let \( x = 3^{1/3} \). This means that \( 3^{2/3} = x^2 \). Substituting \( x \) into the denominator, we have:

\[
1 - x + x^2
\]

So, we can rewrite the original expression as:

\[
\frac{4}{1 - x + x^2}
\]

Next, we will simplify \( 1 - x + x^2 \). We can factor this quadratic expression. To factor \( 1 - x + x^2 \), we can observe that the discriminant may help us identify if it can be factored nicely:

The discriminant \( D \) of the quadratic \( ax^2 + bx + c \) is given by

\[
D = b^2 - 4ac
\]

Here, \( a = 1 \), \( b = -1 \), and \( c = 1 \):

\[
D = (-1)^2 - 4(1)(1) = 1 - 4 = -3
\]

Since the discriminant is negative, \( 1 - x + x^2 \) cannot be factored into real linear factors. We can complete the square:

\[
1 - x + x^2 = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}
\]

Now the denominator can be expressed as:

\[
1 - x + x^2 = \left( x - \frac{1}{2} \right)^2 + \frac{3}{4}
\]

Next, we need to put this back into the fraction:

\[
\frac{4}{\left( x - \frac{1}{2} \right)^2 + \frac{3}{4}}
\]

Since we have:

\[
x = 3^{1/3}
\]

We can note that the original expression doesn't seem to simplify much further. A better approach would be just to leave it as:

\[
\frac{4}{1 - 3^{1/3} + 3^{2/3}}
\]

However, we can confirm that the simplified rational form given the transformations we have made will be indeed:

\[
\frac{4}{1 - 3^{1/3} + 3^{2/3}}
\]

If you want an approximate numerical evaluation, \( 3^{1/3} \approx 1.442 \), and hence:

\[
1 - 3^{1/3} + 3^{2/3} \approx 1 - 1.442 + 2.08 \approx 1.638
\]

Thus, the approximate value would yield:

\[
\frac{4}{1.638} \approx 2.44
\]

But for algebraic simplification, we stick to:

\[
\frac{4}{1 - 3^{1/3} + 3^{2/3}}
\]

is a simplified expression.