If three cubes of volume 512 cm 3 each are joined end to end, find the surface area of the resulting cuboid.

Correct — d. Recover the edge first, then build the cuboid.

Volume of one cube = a³ = 512, so a = ∛512 = 8 cm.

Three joined end to end give a cuboid 24 × 8 × 8 cm — the length triples, the cross-section does not change.

Surface area = 2(lb + bh + hl)
= 2(24×8 + 8×8 + 8×24)
= 2(192 + 64 + 192)
= 2 × 448 = 896 → option (d).

Why the others are wrong

• (a)
  986 cm — 986 is 896 with its first two digits swapped. The bracket totals 192 + 64 + 192 = 448, and twice 448 is 896.
• (b)
  895 cm — 895 is odd. With whole-number edges 2(lb + bh + hl) is always even, and the bracket here comes to 448.
• (c)
  869 cm — 869 swaps the last two digits of 896. Check the bracket: the two long faces are 24 × 8 = 192 each and the two ends are 8 × 8 = 64.

Two ideas meet here. The cube root recovers the edge, and joining cubes in a line changes only one dimension.

512 = 8³, so each edge is 8 cm. Laid end to end, three of them make a 24 × 8 × 8 cuboid: 8 becomes 24 in one direction and stays 8 in the other two.

A second route checks the first. Three separate cubes carry 3 × 6 × 64 = 1152 cm² of surface. Each join buries two faces, and there are two joins, so 1152 − 4 × 64 = 896 cm².

Key facts

• 512 = 8³, so a cube of volume 512 cm³ has an edge of 8 cm.
• Three equal cubes joined end to end form a 24 × 8 × 8 cm cuboid.
• Surface area = 2(lb + bh + hl) = 2(192 + 64 + 192) = 896 cm².
• Joining n cubes in a row buries 2(n − 1) faces, here four faces of 64 cm² each.

Common traps

• Tripling one cube's surface area and forgetting the buried faces
• Cubing 8 instead of taking the cube root of 512
• Treating the join as stretching all three dimensions rather than the length alone

No directly related past PYQ was found.