Answer :

Certainly! Let's solve the expression step by step:

Given:
[tex]\[ (9 \sqrt{8 x} - 1)(8 \sqrt{9 x} + 3) \][/tex]

First, let's use the distributive property to expand the product of the two binomials.

Step 1: Distribute [tex]\(9 \sqrt{8 x}\)[/tex] across [tex]\(8 \sqrt{9 x} + 3\)[/tex]:

[tex]\[ 9 \sqrt{8 x} \cdot 8 \sqrt{9 x} + 9 \sqrt{8 x} \cdot 3 \][/tex]

Step 2: Distribute [tex]\(-1\)[/tex] across [tex]\(8 \sqrt{9 x} + 3\)[/tex]:

[tex]\[ -1 \cdot 8 \sqrt{9 x} - 1 \cdot 3 \][/tex]

Putting these together, we have:

[tex]\[ (9 \sqrt{8 x})(8 \sqrt{9 x}) + (9 \sqrt{8 x})(3) + (-1)(8 \sqrt{9 x}) + (-1)(3) \][/tex]

Let's simplify each term separately.

Term 1:
[tex]\[ (9 \sqrt{8 x})(8 \sqrt{9 x}) = 9 \cdot 8 \cdot \sqrt{8 x} \cdot \sqrt{9 x} = 72 \cdot \sqrt{72 x^2} \][/tex]

Now, [tex]\(\sqrt{72 x^2}\)[/tex] can be simplified further:

[tex]\[ \sqrt{72 x^2} = \sqrt{36 \cdot 2 \cdot x^2} = 6 \sqrt{2} \cdot x \][/tex]

So the first term becomes:
[tex]\[ 72 \cdot 6 \sqrt{2} \cdot x = 432 \sqrt{2} \cdot x \][/tex]

Term 2:
[tex]\[ (9 \sqrt{8 x})(3) = 27 \sqrt{8 x} \][/tex]

We know that [tex]\(\sqrt{8} = 2 \sqrt{2}\)[/tex], therefore:

[tex]\[ 27 \sqrt{8 x} = 27 \cdot 2 \sqrt{2 x} = 54 \sqrt{2} \sqrt{x} = 54 \sqrt{2 x} \][/tex]

Term 3:
[tex]\[ -1 \cdot 8 \sqrt{9 x} = -8 \sqrt{9 x} = -8 \cdot 3 \sqrt{x} = -24 \sqrt{x} \][/tex]

Term 4:
[tex]\[ -1 \cdot 3 = -3 \][/tex]

Putting all terms together, we get:

[tex]\[ 432 \sqrt{2} x + 54 \sqrt{2 x} - 24 \sqrt{x} - 3 \][/tex]

Since [tex]\(\sqrt{2 x} = \sqrt{2} \sqrt{x}\)[/tex], we can rewrite [tex]\(54 \sqrt{2 x}\)[/tex] as [tex]\(54 \sqrt{2} \sqrt{x}\)[/tex].

Combining everything, we obtain:

[tex]\[ 432 \sqrt{2} x + 54 \sqrt{2} \sqrt{x} - 24 \sqrt{x} - 3 \][/tex]

So the resulting expanded form is:

[tex]\[ -24\sqrt{x} + 54\sqrt{2}\sqrt{x} + 432\sqrt{2} x - 3 \][/tex]

That completes the detailed, step-by-step expansion of the given expression.