# Magnified cube

If the lengths of the edges of the cube are extended by 5 cm, its volume will increase by 485 cm3. Determine the surface of both the original and the magnified cube.

Result

S1 =  54 cm2
S2 =  384 cm2

#### Solution:

$(a+5)^3=a^3+485 \ \\ a^3 + 15 \ a^2 + 75 \ a + 125=a^3+485=3^3+485=488 \ \\ \ \\ \ \\ 15 a^2 + 75 a + 125=485 \ \\ \ \\ 15 \ a^2 + 75 \ a + 125=485 \ \\ 15a^2 +75a -360=0 \ \\ \ \\ p=15; q=75; r=-360 \ \\ D=q^2 - 4pr=75^2 - 4\cdot 15 \cdot (-360)=27225 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ -75 \pm \sqrt{ 27225 } }{ 30 } \ \\ a_{1,2}=\dfrac{ -75 \pm 165 }{ 30 } \ \\ a_{1,2}=-2.5 \pm 5.5 \ \\ a_{1}=3 \ \\ a_{2}=-8 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 15 (a -3) (a +8)=0 \ \\ a=a_{1}=3 \ \\ \ \\ S_{1}=6 \cdot \ a^2=6 \cdot \ 3^2=54 \ \text{cm}^2$

Checkout calculation with our calculator of quadratic equations.

$S_{2}=6 \cdot \ (a+5)^2=6 \cdot \ (3+5)^2=384 \ \text{cm}^2$

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