The value of `(sqrt3 + sqrt2)^3 + (sqrt3 -sqrt2)^3 ` is ………..

Answer

To solve the expression \((\sqrt{3} + \sqrt{2})^3 + (\sqrt{3} - \sqrt{2})^3\), we can use the formula for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Here, let:
- \(a = \sqrt{3} + \sqrt{2}\)
- \(b = \sqrt{3} - \sqrt{2}\)

### Step 1: Calculate \(a + b\)

\[ a + b = (\sqrt{3} + \sqrt{2}) + (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} \]

**Hint:** Combine like terms to simplify the expression.

### Step 2: Calculate \(a^2\) and \(b^2\)

\[ a^2 = (\sqrt{3} + \sqrt{2})^2 = 3 + 2 + 2\sqrt{6} = 5 + 2\sqrt{6} \]

\[ b^2 = (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6} \]

**Hint:** Use the formula \((x+y)^2 = x^2 + y^2 + 2xy\) to expand the squares.

### Step 3: Calculate \(ab\)

\[ ab = (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1 \]

**Hint:** Recognize that this is a difference of squares.

### Step 4: Substitute into the sum of cubes formula

Now we can substitute \(a + b\), \(a^2\), \(b^2\), and \(ab\) into the formula:

\[ a^2 + b^2 = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10 \]

So we have:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) = (2\sqrt{3})(10 - 1) = (2\sqrt{3})(9) = 18\sqrt{3} \]

### Final Answer

Thus, the value of \((\sqrt{3} + \sqrt{2})^3 + (\sqrt{3} - \sqrt{2})^3\) is:

\[ \boxed{18\sqrt{3}} \]

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