I2.2 The surface area of a cube is $b \mathrm{~cm}^{2}$. If the length of each side is increased by $3 \mathrm{~cm}$, its volume is increased by $(2 b-a) \mathrm{cm}^{3}$, find the value of $b$.
Let the original length of each side be $x \mathrm{~cm}$.
Old surface area $b \mathrm{~cm}^{2}=6 x^{2} \mathrm{~cm}^{2}$
Original volume $=x^{3} \mathrm{~cm}^{3}$
New length of side $=(x+3) \mathrm{cm}$.
New volume $=(x+3)^{3} \mathrm{~cm}^{3}$
Increase in volume $=\left[(x+3)^{3}-x^{3}\right] \mathrm{cm}^{3}=(2 b-a) \mathrm{cm}^{3}$ $9 x^{2}+27 x+27=2\left(6 x^{2}\right)-3$
$3 x^{2}-27 x-30=0$
$x^{2}-9 x-10=0$
$(x-10)(x+1)=0$
$x=10$
$b=6 x^{2}=600$