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Hint: Let any variable be the side of the cube and the surface area and volume of the cube is 6 times side square and side cube respectively so use these concepts to reach the solution of the question.

Complete step-by-step answer:

As we know that the cube has 6 sides and each side represents the square.

And the area of square is $ = {\left( {{\text{side}}} \right)^2}$

Let the side length of the cube be a unit.

So the surface area (S.A) of the cube is $ = 6{a^2}$ sq. unit

$ \Rightarrow S.A = 6{a^2}$ sq. unit ………………….. (1)

And we all know that the volume (V) of the cube is $ = {\left( {{\text{Side}}} \right)^3}$

$ \Rightarrow V = {a^3}$ unit cube………………………… (2)

Now it is given that the edge of the cube is doubled.

So now the edge of the cube becomes 2a.

So the new surface area ($S.{A_1}$) of the cube is $ = 6{\left( {2a} \right)^2} = 6\left( {4{a^2}}

\right) = 24{a^2}$ sq. unit.

$ \Rightarrow S.{A_1} = 4\left( {6{a^2}} \right)$

Now from equation (1) we have,

$ \Rightarrow S.{A_1} = 4\left( {S.A} \right)$

So the new surface area of the cube is four times the old surface area.

And the new volume (${V_1}$) of the cube is $ = {\left( {2a} \right)^3} = 8{a^3}$

$ \Rightarrow {V_1} = 8{a^3}$

Now from equation (2) we have,

$ \Rightarrow {V_1} = 8\left( V \right)$

So, the new volume of the cube is eight times the old volume.

So, this is the required answer.

Note: Whenever we face such types of problem the key concept is formula of surface area and volume of cube which is stated above then make the side double as given in problem statement and calculate new surface area and volume of the cube as above then convert new surface area and volume in terms of old surface area and volume of the cube as above so, doing this we can easily calculate how much times the surface area and volume of the cube is increased if the edge of the cube is doubled.

Complete step-by-step answer:

As we know that the cube has 6 sides and each side represents the square.

And the area of square is $ = {\left( {{\text{side}}} \right)^2}$

Let the side length of the cube be a unit.

So the surface area (S.A) of the cube is $ = 6{a^2}$ sq. unit

$ \Rightarrow S.A = 6{a^2}$ sq. unit ………………….. (1)

And we all know that the volume (V) of the cube is $ = {\left( {{\text{Side}}} \right)^3}$

$ \Rightarrow V = {a^3}$ unit cube………………………… (2)

Now it is given that the edge of the cube is doubled.

So now the edge of the cube becomes 2a.

So the new surface area ($S.{A_1}$) of the cube is $ = 6{\left( {2a} \right)^2} = 6\left( {4{a^2}}

\right) = 24{a^2}$ sq. unit.

$ \Rightarrow S.{A_1} = 4\left( {6{a^2}} \right)$

Now from equation (1) we have,

$ \Rightarrow S.{A_1} = 4\left( {S.A} \right)$

So the new surface area of the cube is four times the old surface area.

And the new volume (${V_1}$) of the cube is $ = {\left( {2a} \right)^3} = 8{a^3}$

$ \Rightarrow {V_1} = 8{a^3}$

Now from equation (2) we have,

$ \Rightarrow {V_1} = 8\left( V \right)$

So, the new volume of the cube is eight times the old volume.

So, this is the required answer.

Note: Whenever we face such types of problem the key concept is formula of surface area and volume of cube which is stated above then make the side double as given in problem statement and calculate new surface area and volume of the cube as above then convert new surface area and volume in terms of old surface area and volume of the cube as above so, doing this we can easily calculate how much times the surface area and volume of the cube is increased if the edge of the cube is doubled.

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