How to solve a 5×5 cube
Solve the six centres, combine each set of three edge pieces and reduce the puzzle to a normal 3×3.
How the reduction method works
A 5×5 initially has many separate centre and edge pieces. Reduction groups these pieces together until the puzzle behaves like a 3×3.
Solve the six centres
As on a 4×4, begin by building centre bars. The 5×5 is more convenient because its fixed middle centres show exactly which colour belongs on every face.
Build the middle bar
Choose a fixed centre and join the two matching centre pieces on either side of it. This creates the middle bar of your first 3×3 centre block.
The fixed centres remove the possibility of building the colour scheme incorrectly: simply build each centre around its fixed middle piece.
Build another bar
Use the upper and inner slices to join three more white pieces without breaking the middle bar.
Complete the first centre
Insert the second bar alongside the middle bar, then build and insert the final bar in the same way.
Start the opposite colour
Rotate the cube so the first centre is on the bottom. Begin building the opposite centre around its fixed middle piece. Because one centre is already solved, insert pieces with slice–turn–slice sequences.
2R U 2R'Insert complete bars
Once the middle bar is complete, build another 1×3 bar in the working layer. Insert it with a double upper turn so the solved centre is restored when the inner slice returns.
2R U2 2R'Preserve one colour while finishing the other
Build a middle bar on the front. Build another bar for the front colour on the upper face, then bring it down while preserving the completed centres with 2R' U2 2R.
This finishing sequence is not essential to memorise immediately, but becomes increasingly useful on larger cubes:
2R U' M' U 2R' U' MShow move-by-move notation images
Pair the twelve edges
Every 5×5 edge contains one fixed middle edge and two movable outer edge pieces, often called wings. Pair both wings with their matching middle edge to create one 3×1 edge triplet.
Add one wing at a time
Choose a middle edge and find one matching wing. If the lower colour of the wing matches the middle edge on the front, bring it into position with 2L'.
Move the completed two-piece edge into a free storage slot on the left or right:
U L U' or U' R' UShow notation images
Repeat until all eight storage slots—four on each side—contain an edge with two correctly joined pieces.
Complete the stored edge triplets
Move an edge with two solved pieces into the working layer, then find its remaining wing. Insert the wing so restoring the inner slice does not destroy another paired edge.
In this example:
2L' L U L' U' 2LShow move-by-move notation images
Repeat until the eight left and right storage positions contain complete edge triplets.
Solve the final four edges
The last four edges cannot be stored in empty side slots, so finish them using flips and cycles. Inspect which pieces need to exchange, open the appropriate slice, replace a wing or middle edge and restore the slice.
Flip an incorrectly arranged edge
The fixed middle edge makes the target easier to recognise, but the sequence follows the same slice–flip–slice idea:
2L' U L U' L F' L' F 2LShow notation images
Move the fixed middle-edge piece
When the middle piece must move, cycle it through the top-back position:
M' U2 MRecognise and solve parity
Repeated flips and cycles will solve the final four edges—or leave this parity case. Unlike on a 4×4, the fixed middle edge makes it possible to recognise the problem before reaching OLL.
The algorithm is the same 4×4 OLL-parity algorithm:
Rw2 B2 U2 Lw U2 Rw' U2 Rw U2 F2 Rw F2 Lw' B2 Rw2Show move-by-move notation images
Finish the cube as a 3×3
Once all six centres and twelve edge triplets are complete, treat every 3×3 centre block as one centre and every edge triplet as one edge.
Solve the reduced cube using your normal 3×3 method. There are no additional PLL-parity cases on a 5×5: once edge-pairing parity has been handled, the final stage contains only ordinary 3×3 cases.
