Explained: "If 3(x3−1x3)3=2, then x3+1x3= what?"

Dillan Gibbs

Answered question

2022-02-07

\sqrt[3]{3 (\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}})} = 2

, then

\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} =

what?

ithangesf4

Beginner2022-02-08Added 16 answers

Step 1 We start with the original function:

\sqrt[3]{3 (\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}})} = 2

and we want to solve for:

\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}

(note the + instead of -) which means we need to do a bit more work and solve for x and then use our x value to solve for:

(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}})

We then take each side to the exponent 3 (which gets rid of the cube root on the left hand side):

(\sqrt[3]{3 (\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}})})^{3} = 2^{3}

Step 2 Taking the cube root of a number is equivalent to taking a fractional exponent of 1/3 and so this simplifies to:

(3 (\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}))^{\frac{3}{3}} = 8

3 (\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}) = 8

We then divide both sides by 3 (to get ride of the multiplication by 3 on the left hand side)

3 \frac{\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}}}{3} = \frac{8}{3}

\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} = \frac{8}{3}

Next we multiply both sides by

\sqrt[3]{x}

to remove it from the denominator of the second term

{\sqrt[3]{x}}^{2} - 1 = \frac{8 \cdot \sqrt[3]{x}}{3}

{\sqrt[3]{x}}^{2} - \frac{8 \cdot \sqrt[3]{x}}{3} - 1 = 0

You might not immediately recognize it, but this is a quadratic equation. To make that more obvious, were

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