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Answer :
We start with a box whose original dimensions are given by
[tex]$$
x,\quad 2x,\quad \text{and} \quad 3x.
$$[/tex]
After increasing each dimension by 2, the new dimensions become
[tex]$$
x + 2,\quad 2x + 2,\quad \text{and} \quad 3x + 2.
$$[/tex]
The volume [tex]$V$[/tex] of the box is the product of its three dimensions. Therefore, we have
[tex]$$
V = (x + 2)(2x + 2)(3x + 2).
$$[/tex]
Let's multiply these factors step by step.
Step 1. Multiply the first two factors:
[tex]\[
(x + 2)(2x + 2).
\][/tex]
We can use the distributive property (FOIL method):
[tex]\[
(x + 2)(2x) + (x + 2)(2) = 2x(x + 2) + 2(x + 2).
\][/tex]
Factor out the common factor [tex]$(x+2)$[/tex]:
[tex]\[
= (x + 2)(2x + 2).
\][/tex]
Carrying out the multiplication:
[tex]\[
= 2x^2 + 4x + 2x + 4 = 2x^2 + 6x + 4.
\][/tex]
Step 2. Now, multiply the result by the third factor [tex]$(3x + 2)$[/tex]:
[tex]\[
(2x^2 + 6x + 4)(3x + 2).
\][/tex]
Use the distributive property to multiply each term in [tex]$2x^2 + 6x + 4$[/tex] by each term in [tex]$3x + 2$[/tex]:
[tex]\[
2x^2 \cdot 3x + 2x^2 \cdot 2 + 6x \cdot 3x + 6x \cdot 2 + 4 \cdot 3x + 4 \cdot 2.
\][/tex]
Calculate each product:
[tex]\[
= 6x^3 + 4x^2 + 18x^2 + 12x + 12x + 8.
\][/tex]
Combine like terms:
[tex]\[
6x^3 + (4x^2 + 18x^2) + (12x + 12x) + 8 = 6x^3 + 22x^2 + 24x + 8.
\][/tex]
Thus, the volume of the box is
[tex]$$
\boxed{6x^3 + 22x^2 + 24x + 8}.
$$[/tex]
[tex]$$
x,\quad 2x,\quad \text{and} \quad 3x.
$$[/tex]
After increasing each dimension by 2, the new dimensions become
[tex]$$
x + 2,\quad 2x + 2,\quad \text{and} \quad 3x + 2.
$$[/tex]
The volume [tex]$V$[/tex] of the box is the product of its three dimensions. Therefore, we have
[tex]$$
V = (x + 2)(2x + 2)(3x + 2).
$$[/tex]
Let's multiply these factors step by step.
Step 1. Multiply the first two factors:
[tex]\[
(x + 2)(2x + 2).
\][/tex]
We can use the distributive property (FOIL method):
[tex]\[
(x + 2)(2x) + (x + 2)(2) = 2x(x + 2) + 2(x + 2).
\][/tex]
Factor out the common factor [tex]$(x+2)$[/tex]:
[tex]\[
= (x + 2)(2x + 2).
\][/tex]
Carrying out the multiplication:
[tex]\[
= 2x^2 + 4x + 2x + 4 = 2x^2 + 6x + 4.
\][/tex]
Step 2. Now, multiply the result by the third factor [tex]$(3x + 2)$[/tex]:
[tex]\[
(2x^2 + 6x + 4)(3x + 2).
\][/tex]
Use the distributive property to multiply each term in [tex]$2x^2 + 6x + 4$[/tex] by each term in [tex]$3x + 2$[/tex]:
[tex]\[
2x^2 \cdot 3x + 2x^2 \cdot 2 + 6x \cdot 3x + 6x \cdot 2 + 4 \cdot 3x + 4 \cdot 2.
\][/tex]
Calculate each product:
[tex]\[
= 6x^3 + 4x^2 + 18x^2 + 12x + 12x + 8.
\][/tex]
Combine like terms:
[tex]\[
6x^3 + (4x^2 + 18x^2) + (12x + 12x) + 8 = 6x^3 + 22x^2 + 24x + 8.
\][/tex]
Thus, the volume of the box is
[tex]$$
\boxed{6x^3 + 22x^2 + 24x + 8}.
$$[/tex]
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