Answer:
[tex]800[/tex]
Step-by-step explanation:
[tex]\text{Solution:}\\\text{Let the three numbers be }x,\ y\text{ and }z.\\[/tex]
[tex]\text{1st case:}\\x+y+z=14.....(1)[/tex]
[tex]\text{2nd case:}\\x^2+y^2+z^2=100.....(2)[/tex]
[tex]\text{3rd case:}\\xyz=24[/tex]
[tex]\bold{Step\ 1: }\text{ Cube equation(1).}[/tex]
[tex](x+y+z)^3=14^3[/tex]
[tex]\bold{Step\ 2:}\text{ Expand}[/tex]
[tex]\text{or, }(x+y)^3+3(x+y)^2z+3(x+y)z^2+z^3=2744\\[/tex]
[tex]\text{or, }x^3+y^3+3x^2y+3xy^2+3z(x^2+2xy+y^2)+3z^2(x+y)+z^3=2744[/tex]
[tex]\text{or, }x^3+y^3+3x^2y+3xy^2+3x^2z+6xyz+3y^2z+3xz^2+3yz^2+z^3=2744[/tex]
[tex]\bold{Step\ 3:}\text{ Group the terms }x^3,\ y^3\text{ and }z^3\text{ and the terms with }x^2,\ y^2\text{ and }z^2\text{ at}\\\text{one place.}[/tex]
[tex]\text{or, }x^3+y^3+z^3+3x^2y+3x^2z+3xy^2+3y^2z+3xz^2+3yz^2+6xyz=2744[/tex]
[tex]\bold{Step\ 4:}\text{ Factor out }3x^2,\ 3y^2\text{ and }3z^2\text{ from the grouped terms. }xyz\text{ can be }\\\text{written as 24 from equation(3).}[/tex]
[tex]\text{or, }x^3+y^3+z^3+3x^2(y+z)+3y^2(x+z)+3z^2(x+y)+6(24)=2744[/tex]
[tex]\bold{Step\ 5: }\text{ Take the help of equation(1) to replace the values }(y+z),\ (x+z)\\\text{ and }(x+y).\text{ Then expand again.}[/tex]
[tex]\text{or, }x^3+y^3+z^3+3x^2(14-x)+3y^2(14-y)+3z^2(14-z)+144=2744[/tex]
[tex]\text{or, }x^3+y^3+z^3+42x^2-3x^3+42y^2-3y^3+42z^2-3z^3=2600\\[/tex]
[tex]\bold{Step\ 6:}\text{ Subtract the like terms and group the terms }42x^2,\ 42y^2\text{ and }42z^2.[/tex]
[tex]\text{or, }-2x^3-2y^2-2z^3+42x^2+42y^2+42z^2=2600[/tex]
[tex]\text{or, }-2(x^3+y^3+z^3)+42(x^2+y^2+z^2)=2600[/tex]
[tex]\bold{Step\ 7:}\ x^2+y^2+z^2=100\text{ from equation(2), so replace this value. Then }\\\text{simplify further to get the value of }x^3+y^3+z^3.[/tex]
[tex]\text{or, }-2(x^3+y^3+z^3)+4200=2600\\[/tex]
[tex]\text{or, }-2(x^3+y^3+z^3)=-1600\\[/tex]
[tex]\text{or, }x^3+y^3+z^3=800[/tex]
[tex]\text{Therefore, the sum of cubes of those numbers is 800.}[/tex]
