We start with the given expression:
[tex]\[ x = p^{\frac{1}{3}} - p^{-\frac{1}{3}} \][/tex]
We need to show that:
[tex]\[ x^3 + 3x = p - \frac{1}{p} \][/tex]
First, let's calculate \(x^3\):
[tex]\[ x = p^{\frac{1}{3}} - p^{-\frac{1}{3}} \][/tex]
Now consider the expression \(x^3\):
[tex]\[ x^3 = (p^{\frac{1}{3}} - p^{-\frac{1}{3}})^3 \][/tex]
Using the binomial expansion for the cube of a binomial, we get:
[tex]\[ x^3 = \left(p^{\frac{1}{3}}\right)^3 - 3 \left(p^{\frac{1}{3}}\right)^2 \left(p^{-\frac{1}{3}}\right) + 3 \left(p^{\frac{1}{3}}\right) \left(p^{-\frac{1}{3}}\right)^2 - \left(p^{-\frac{1}{3}}\right)^3 \][/tex]
This simplifies to:
[tex]\[ x^3 = p - 3 p^{\frac{1}{3}} p^{-\frac{1}{3}} + 3 p^{\frac{1}{3}} p^{-\frac{1}{3}} - \frac{1}{p} \][/tex]
Notice that:
[tex]\[ p^{\frac{1}{3}} p^{-\frac{1}{3}} = p^{\frac{1}{3} - \frac{1}{3}} = p^0 = 1 \][/tex]
So the expression becomes:
[tex]\[ x^3 = p - 3(1) + 3(1) - \frac{1}{p} \][/tex]
[tex]\[ x^3 = p - \frac{1}{p} \][/tex]
Now, to complete the proof, add \(3x\), where \(x = p^{\frac{1}{3}} - p^{-\frac{1}{3}}\):
[tex]\[ x^3 + 3x = \left(p - \frac{1}{p}\right) + 3\left(p^{\frac{1}{3}} - p^{-\frac{1}{3}}\right) \][/tex]
Substituting back, since \(p^{\frac{1}{3}} = a\) and \(p^{-\frac{1}{3}} = \frac{1}{a}\), we simplify to:
[tex]\[ x^3 = -\frac{1}{p} + p \][/tex]
Thus, we reach the final conclusion:
[tex]\[ x^3 + 3x = p - \frac{1}{p} \][/tex]
Therefore, the expression is verified to be correct.
