Let the radius of the sphere be r cm.
`:.` the volume of th sphere `=(4)/(3)pir^(3)` cc.
If the radiuys of the sphere be incereased by 3 cm, then its volume will be `(4)/(3) pi(r+3)^(3)` cc.
As per question `(4)/(3)pi (r+3)^(3)- (4)/(3)pir^(3)=264`.
or, `(4)/(3)pi[(r+3)^(3)-r^(3)]=264 "or" (4)/(3)xx(22)/(7)(r^(3)+9r^(2)+27r+27-r^(3))=264`
or, `(4)/(3)xx(22)/(7)(9r^(2)+27r+27)=264 "or," 9r^(2)+27r+27=(264xx3xx7)/(4xx22)`
or,`9(r^(2)+3r+3)=63 "or,"r^(2)+3r+3=7 "or" r^(2)+3r+3-7=0 "or" r^(2)+3r-4=0`
or, `r^(2)`+4r-r-4=0 or, r(r+4)-1(r+4)=0 or, (r+4) (r-1)=0
`:.` either r+4=0, or, r-1 =0 `rArr` r=-4 or, r=1
But the value of r cannot be negative, `:.` r=1. Hence the requied radius of the sphere was 1 cm.
