The total area of a cube is expressed by A(x) = 24x^2+24x+6. What is the volume of this cube?

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Answer 1 · 2018-03-02T04:12:40

$8 x^{3} + 12 x^{2} + 6 x + 1$ $8x^3+12x^2+6x+1$

I'm going to assume you meant the surface area is given by $A (x)$ $A(x)$ .

We have $A (x) = 24 x^{2} + 24 x + 6$ $A(x)=24x^2+24x+6$

The formula for the surface area of a cube is given by $6 k^{2}$ $6k^2$ , where $k$ $k$ is the length of a side.

We can say that:

$6 k^{2} = 24 x^{2} + 24 x + 6$ $6k^2=24x^2+24x+6$

$k^{2} = 4 x^{2} + 4 x + 1$ $k^2=4x^2+4x+1$

$k^{2} = {(2 x + 1)}^{2}$ $k^2=(2x+1)^2$

$k = 2 x + 1$ $k=2x+1$

So the length of a side is $2 x + 1$ $2x+1$ .

On the other hand, $V (x)$ $V(x)$ , the volume of he cube, is given by $k^{3}$ $k^3$ .

Here, $k = 2 x + 1$ $k=2x+1$

So we can say:

$V (x) = k^{3} = {(2 x + 1)}^{3}$ $V(x)=k^3=(2x+1)^3$

$V (x) = {(2 x + 1)}^{2} (2 x + 1)$ $V(x)=(2x+1)^2(2x+1)$

$V (x) = (2 x + 1) (4 x^{2} + 4 x + 1)$ $V(x)=(2x+1)(4x^2+4x+1)$

$V (x) = 8 x^{3} + 12 x^{2} + 6 x + 1$ $V(x)=8x^3+12x^2+6x+1$

So the volume of this cube is given by $8 x^{3} + 12 x^{2} + 6 x + 1$ $8x^3+12x^2+6x+1$