# Math Help - When each edge of a cube is decreased by 1 cm, its volume is decreased by 91 cm^3.

1. ## When each edge of a cube is decreased by 1 cm, its volume is decreased by 91 cm^3.

Hello! I would like some help with this task.

When each edge of a cube is decreased by 1 cm, its volume is decreased by 91 cm^3. Find the length of a side of the orginal cube.

Thank you!

2. $V_1 = a^3$ and $V_2 = (a-1)^3$

$V_2 = V_1 - 91$ so $V_1 - 91 = a^3 - (a-1)^3$

edit: $a^3$ will cancel but you'll need the binomial expansion to expand $(a-1)^3$

3. I do not understand this V1 - 91 = a^3 - (a-1)^3 can you please explain?

4. Originally Posted by Anna55
I do not understand this V1 - 91 = a^3 - (a-1)^3 can you please explain?
Hi Anna55,

The first volume is $V_1=a^3$

The volume after side reduction is $V_2=(a-1)^3$

The difference between the two volumes is $a^3-(a-1)^3=91$

Expand and solve. Throw out the negative result.

5. Originally Posted by Anna55
I do not understand this V1 - 91 = a^3 - (a-1)^3 can you please explain?
It means just what it says- cube a and a-1 and subtract. Do you know how to multiply a- 1 by itself? However, the "V1- 91" is incorrec- perhaps that is what is confusing you. It should be a^3- (a-1)^3= 91. Multiply that out and solve for a. As e^(i pu) said, the a^4 terms will cancel leaving a quadratic equation to solve.

If you do not know how to multiply (a- 1)^3, get back to us.

6. Thank you very much HallsofIvy and masters , I understand everything now.

7. Originally Posted by Anna55
Hello! I would like some help with this task.

When each edge of a cube is decreased by 1 cm, its volume is decreased by 91 cm^3. Find the length of a side of the orginal cube.

Thank you!
hi Anna,

you can also calculate this with a quadratic equation.

Suppose all the sides of the cube is x.
Now, cut off a vertical slice of 1 unit deep (say from the right-hand side of the cube).

The volume of the part cut off is $x(x)(1)=x^2$ cubic units.

Next, cut off a piece from the front or back (say from the back).
One side length is now (x-1) units long,
therefore the volume of the piece cut off is $x(x-1)(1)=x^2-x$ cubic units.

Finally cut off a piece from the top.
This time, 2 sides have length (x-1), so the volume of the piece cut off is

$(1)(x-1)^2=x^2-2x+1$ cubic units.

The total volume of the removed parts is $x^2+x^2-x+x^2-2x+1=3x^2-3x+1$

Hence

$3x^2-3x+1=91\Rightarrow\ 3x^2-3x-90=0\Rightarrow\ x^2-x-30=0$

$\Rightarrow\ (x-6)(x+5)=0\Rightarrow\ x=6\;cm$

You could of course try a shortcut, which works only if the side length is an integer.

$2^3-1^3=8-1=7$

$3^3-2^3=27-8=19$

$4^3-3^3=64-27=37$

$5^3-4^3=125-64=61$

and so on.