Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

If [tex]$x = p^{\frac{1}{3}} - p^{-\frac{1}{3}}[tex]$[/tex], show that [tex]$[/tex]x^3 + 3x = p - \frac{1}{p}$[/tex].

Sagot :

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.