Question

A small cube with side length 6y is placed inside a larger cube with side length 4/2. What is the difference in the
volume of the cubes?
• (4x2 -by) (16** +24x2y+38y?)
• (63-4x2) 36y2 +24x+y+ 16x
• (4x2-by)(4x4 +24x+y+by?)
• (4x2 + 8y)(18** – 24x+y+38y?)

Answer 1

Answer:

Step-by-step explanation:

Side length of a small cube = 6y

Side length of a larger cube = 4x²

This small cube is placed inside a larger cube. We need to find the difference in the  volume of the cubes.

Volume of small cube, $V_1=(6y)^3=216y^3$

Volume of lager cube,

$V_2=(4x^2)^3\\\\=64x^6$

Difference in the volume of the cubes,

$\Delta V=V_2-V_1\\\\=64x^6-216y^3$

We can simplify the above expression :

$\because \ (a^3-b^3)=(a-b)(a^2+ab+b^2)\\\\(4x^2)^3-(6y)^3=(4x^2-6y)((4x^2)^2+4x^2\times 6y +(6y)^2)\\\\=(4x^2-6y)(16x^4+24x^2y+36y^2)$

So, the difference in the volumes of the cubes is $(4x^2-6y)(16x^4+24x^2y+36y^2)$.

Answer 2

Answer:

see image below

Step-by-step explanation: