Answer:

To find the value of x³ + 3x + 3 we can start by substituting the given value of x.

Let's begin by simplifying (√2 + 1)^(1/3):

Let a = √2 + 1

(a)^3 = (√2 + 1)^3

a^3 = 2√2 + 3√2 + 3 + 1

a^3 = 3√2 + 4 + 3

a^3 = 3√2 + 7

Now let's simplify (√2 - 1)^(1/3):

Let b = √2 - 1

(b)^3 = (√2 - 1)^3

b^3 = 2√2 - 3√2 + 3 - 1

b^3 = -√2 + 2

We can rewrite it as:

-b^3 = √2 - 2

Substituting these values back into the expression x = (√2 + 1)^(1/3) - (√2 - 1)^(1/3):

x = a - b

x = 3√2 + 7 - (-√2 + 2)

x = 3√2 + √2 + 7 + 2

x = 4√2 + 9

Now let's substitute this value back into x³ + 3x + 3:

x³ + 3x + 3 = (4√2 + 9)³ + 3(4√2 + 9) + 3

= (4√2)³ + 3(4√2)²(9) + 3(4√2)(9²) + 9³ + 3(4√2)(1) + 3(9) + 3

= 32√2 + 144√2 + 648 + 729 + 12√2 + 27 + 3

= 44√2 + 1416 + 759 + 3

= 44√2 + 2178

Therefore x³ + 3x + 3 = 44√2 + 2178.

Verified answer

Answer:

63

Step-by-step explanation:

To find the value of \(x\) and then \(x^3 + 3x + 3\), we can first simplify \(x\) using the property of a cube root of a difference of cubes, and then proceed to calculate \(x^3 + 3x + 3\).

Given: \(x = (\sqrt{2} + 1)^{\frac{1}{3}} - (\sqrt{2} - 1)^{\frac{1}{3}}\)

Step 1: Simplify \(x\)

Let's use the formula for the difference of cubes, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):

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

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

Simplifying further:

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

Step 2: Calculate \(x^3 + 3x + 3\)

\[x^3 = 8 \left((\sqrt{2} + 1)^2 + 3(\sqrt{2} + 1)(\sqrt{2} - 1) + (\sqrt{2} - 1)^2\right)\]

\[x^3 = 8 \left(2 + 3(\sqrt{2}^2 - 1) + 2\right)\]

\[x^3 = 8 \left(2 + 3(2 - 1) + 2\right)\]

\[x^3 = 8 \times 6 = 48\]

Now, let's calculate \(3x\):

\[3x = 6 \left((\sqrt{2} + 1)^{\frac{2}{3}} + (\sqrt{2} + 1)^{\frac{1}{3}}(\sqrt{2} - 1)^{\frac{1}{3}} + (\sqrt{2} - 1)^{\frac{2}{3}}\right)\]

\[3x = 6 \times 2 = 12\]

Finally, \(x^3 + 3x + 3\) can be calculated as follows:

\[x^3 + 3x + 3 = 48 + 12 + 3 = 63\]

So, the value of \(x^3 + 3x + 3\) is \(63\).