Question

three cubes of iron whose edges are 6cm,8cm,and 10cm are melted and formed into a single cube.The edge of the new cube formed is_

Answer 1

volume of first cube =
${6}^{3} \\ 6 \times 6 \times 6 \\ = (206 \div 100 ) {m}^{3} = 2.06 {m}^{3}$
volume of second cube=
${8}^{3} \\ 8 \times 8 \times 8 \\ = (512 \div 100) {m}^{3} =5.12 {m}^{3}$
volume of third cube=
${10}^{3} \\ 10 \times 10 \times 10 \\ = (1000 \div 100) {m}^{3} = 10 {m}^{3}$
Adding the volumes of all cube =
$2.06 + 5.12 + 10 = 17.18 {m}^{3}$
volume of new cube =
$17.18 = {side}^{3} \\ 17.18 = x \times x \times x \\ 17.18 = 3x \\ 17.18 \div 3 = x \\ 5.73 m = x \\ 5.73 \times 100 = 573cm$
edge of new cube is 573cm