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.


S1 =  54 cm2
S2 =  384 cm2


(a+5)3=a3+485 a3+15 a2+75 a+125=a3+485=33+485=488   15a2+75a+125=485  15 a2+75 a+125=485 15a2+75a360=0  p=15;q=75;r=360 D=q24pr=752415(360)=27225 D>0  a1,2=q±D2p=75±2722530 a1,2=75±16530 a1,2=2.5±5.5 a1=3 a2=8   Factored form of the equation:  15(a3)(a+8)=0 a=a1=3  S1=6 a2=6 32=54 cm2(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

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S2=6 (a+5)2=6 (3+5)2=384 cm2S_{2}=6 \cdot \ (a+5)^2=6 \cdot \ (3+5)^2=384 \ \text{cm}^2

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