A closed aluminum rectangular box has inner dimensions x centimeters

The question itself is very clear and not hard but it took me ages to multiple three brackets without a mistake. Is there a trick?
Thanks

You have to put pen to paper as verbal calculation of the bracket will make issues complicated.
Try solving first two brackets first and then multiply with the third one.
Also, to avoid careless mistakes, always follow a pattern so that you don't miss anything.
for eg ( a + b) ( c + d) = a( c + d) + b ( c + d) always multiply with a pattern .
Hope this helps.

Start with multiplying the first two in the product of the three factors:

\((x+2)(y+2)(z+2)=(xy+2x+2y+4)(z+2)\)

No need to fully carry out the other multiplication. You can see that there will be one term of \(xyz\), which will cancel out in the final expression with \(-xyz\).
In addition, you will have terms containing products of two factors - like \(xy, \,\,xz\), and \(yz\). All will have a coefficient of 2 in front.
Then the terms with \(x, \,\,y,\) and \(z\), all have a coefficient of 4 in front.
Answers A, B, and C can be eliminated. Between D and E, obviously E wins.

A closed aluminum rectangular box has inner dimensions x centimeters by y centimeters by z centimeters. Each of the six sides of the box is 1 centimeter thick. Calculate the volume of the aluminium, in cubic centimeters?

A. xyz + 8
B. 2(xy + xz + yz + 4)
C. 2(xy + xz + yz) – xyz
D. 2(xy + xz + yz + x + y + z + 4)
E. 2(xy + xz + yz + 2x + 2y + 2z + 4)

We are trying to find the volume of the aluminum, NOT the empty space inside the box.

In the INNER dimensions are x*y*z then the dimensions of the box including the one inch thick sides are going to be two CM greater as shown in the attached diagram (in a way, this is similar to a "walkway surrounding a rectangular garden" problem, except in 3 dimensions) Therefore, the length of the box (not just the interior dimensions) is (L+2) The width is (W+2) and the height is (H+2) --> (L+2)*(W+2)*(H+2) is the volume of the box if we consider the volume of the sides included. If we are to figure out the volume of JUST the sides we subtract from (L+2)*(W+2)*(H+2) the volume of the empty space (L*W*H)

(L+2)*(W+2)*(H+2) - (LWH)
(LW+2L+2W+4)*(H+2) - (LWH)
LWH+2LH+2WH+4H+2LW+4L+4W+8 - (LWH)
2LH+2WH+4H+2LW+4L+4W+8
2(LH+WH+LW+2H+2L+2W+4)

Answer E.

Hi - can you please help me understand how you know that based on the 1cm thickness, you should expand each side x/y/z by +2?

Thanks

you certainly know that the xyz will be cancelled out and we will definitely have one 8 in the solution along with one set of double digit variable(such as xy,yz,zx,with what ever coeeficient may be ) and one set of single variable set suchas x,y,z.the suggested set is availble in only E.DONT TOUCH PEN AND PAPER ,BKZ BRAIN IS FASTER THAN HAND.

I am astonished, today I took my test and this was my 2ND question! How is this possible? The solution look quite simple here, but I though arithmetically would take too much time. I solve it with smart numbers (2, 3 and 4), but took me ages to reach the solution:

Inner volume: 2x3x4=24
Exterior volume: 3x4x5-24 = 36.

Then you test all the possible solutions (took me like 4 minutes...)

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