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.

Correct result:

S₁ =  54 cm²
S₂ =  384 cm²

Solution:

$$\begin{gathered} (a+5{)}^3=a^3+485 \\ {a}^3+15a^2+75a+125=a^3+485=3^3+485=488 \\ 15a^2+75a+125=485 \\ 15a^2+75a+125=485 \\ 15a^2+75a-360=0 \\ p=15;q=75;r=-360 \\ D=q^2-4pr=75^2-4\cdot 15\cdot (-360)=27225 \\ D>0 \\ {a}_{1,2}={\displaystyle \frac{-q\pm \sqrt{D}}{2p}}={\displaystyle \frac{-75\pm \sqrt{27225}}{30}} \\ {a}_{1,2}={\displaystyle \frac{-75\pm 165}{30}} \\ {a}_{1,2}=-2.5\pm 5.5 \\ {a}_1=3 \\ {a}_2=-8 \\ \text{ Factored form of the equation: } \\ 15(a-3)(a+8)=0 \\ a=a_1=3 \\ {S}_1=6\cdot a^2=6\cdot 3^2=54{\text{cm}}^2 \end{gathered}$$

Our quadratic equation calculator calculates it.

$S_2=6\cdot (a+5{)}^2=6\cdot (3+5{)}^2=384{\text{cm}}^2$

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