Trong so hoc, lap phuong cua mot so n co nghia la nhan 3 lan gia tri cua no voi nhau:

n3 = n × n × n.
Hay cung co the hieu la lay tich cua no voi binh phuong cua no:

n3 = n × n2.
Đay chinh la cong thuc đe tinh the tich cho mot khoi lap phuong co chieu dai cac canh la n.

Lap phuong la mot ham le:

(−n)3 = −(n3).
Bieu đo cua ham lap phuong f: x → x3 (hoac phuong trinh y = x3) đuoc biet đen nhu la hinh parabe hinh khoi. Boi vi lap phuong la mot ham so le, đuong cong nay co mot điem đoi xung o goc, nhung khong co truc đoi xung.


Lap phuong cua so nguyen Lap phuong cua cac so nguyen tu 0 đen 60 la:(day so A000578 trong bang OEIS):

Noi theo hinh hoc, mot so nguyen duong m la mot so lap phuong hoan hao neu va chi khi nao co the sap xep cac khoi hinh khoi ran thanh mot khoi ran lon hon. Vi du, 27 khoi nho co the đuoc sap xep thanh mot khoi lon hon voi su xuat hien cua mot khoi rubic lap phuong, tu 3 × 3 × 3 = 27.
Su chenh lech giua lap phuong cua cac so nguyen lien tiep co the đuoc bieu dien nhu sau:

n3 − (n − 1)3 = 3(n − 1)n + 1.
hoac

(n + 1)3 − n3 =  3(n + 1)n + 1.
Khong co so am nao la so lap phuong hoan hao, vi lap phuong cua mot so am la so am. Vi du, (−4) × (−4) × (−4) = −64.
Chu so tan cung cua lap phuong so co chu so tan cung la 0-9:
0	1	8	7	4	5	6	3	2	9


Tong cua lap phuong n so đau tien Tong cua lap phuong n so đau tien bang binh phuong cua tong n so đau tien:

  
    
      
        
          1
          
            3
          
        
        +
        
          2
          
            3
          
        
        +
        ⋯
        +
        
          n
          
            3
          
        
        =
        (
        1
        +
        2
        +
        ⋯
        +
        n
        
          )
          
            2
          
        
        =
        
          
            (
            
              
                
                  n
                  (
                  n
                  +
                  1
                  )
                
                2
              
            
            )
          
          
            2
          
        
        =
        (
        
          C
          
            n
            +
            1
          
          
            2
          
        
        
          )
          
            2
          
        
      
    
    {\displaystyle 1^{3}+2^{3}+\dots +n^{3}=(1+2+\dots +n)^{2}=\left({\frac {n(n+1)}{2}}\right)^{2}=(C_{n+1}^{2})^{2}}
  
    (1)
Trong đo, 
  
    
      
        
          C
          
            n
            +
            1
          
          
            2
          
        
      
    
    {\displaystyle C_{n+1}^{2}}
  
 la to hop chap 2 cua n+1.
Cong thuc cua Charles Wheatstone (1854):

  
    
      
        
          n
          
            3
          
        
        =
        
          
            
              
                
                  (
                  
                    
                      n
                      
                        2
                      
                    
                    −
                    n
                    +
                    1
                  
                  )
                
                +
                
                  (
                  
                    
                      n
                      
                        2
                      
                    
                    −
                    n
                    +
                    1
                    +
                    2
                  
                  )
                
                +
                
                  (
                  
                    
                      n
                      
                        2
                      
                    
                    −
                    n
                    +
                    1
                    +
                    4
                  
                  )
                
                +
                ⋯
                +
                
                  (
                  
                    
                      n
                      
                        2
                      
                    
                    +
                    n
                    −
                    1
                  
                  )
                
              
              ⏟
            
          
          
            n
            
               so le lien tiep
            
          
        
        .
      
    
    {\displaystyle n^{3}=\underbrace {\left(n^{2}-n+1\right)+\left(n^{2}-n+1+2\right)+\left(n^{2}-n+1+4\right)+\cdots +\left(n^{2}+n-1\right)} _{n{\text{ so le lien tiep}}}.}
  

Đe chung minh cong thuc (1) chung ta co the dung cach sau:

  
    
      
        
          
            
              
                
                  ∑
                  
                    k
                    =
                    1
                  
                  
                    n
                  
                
                
                  k
                  
                    3
                  
                
              
              
                
                =
                1
                +
                8
                +
                27
                +
                64
                +
                ⋯
                +
                
                  n
                  
                    3
                  
                
              
            
            
              
              
                
                =
                
                  
                    
                      1
                      ⏟
                    
                  
                  
                    
                      1
                      
                        3
                      
                    
                  
                
                +
                
                  
                    
                      
                        3
                        +
                        5
                      
                      ⏟
                    
                  
                  
                    
                      2
                      
                        3
                      
                    
                  
                
                +
                
                  
                    
                      
                        7
                        +
                        9
                        +
                        11
                      
                      ⏟
                    
                  
                  
                    
                      3
                      
                        3
                      
                    
                  
                
                +
                
                  
                    
                      
                        13
                        +
                        15
                        +
                        17
                        +
                        19
                      
                      ⏟
                    
                  
                  
                    
                      4
                      
                        3
                      
                    
                  
                
                +
                ⋯
                +
                
                  
                    
                      
                        
                          (
                          
                            
                              n
                              
                                2
                              
                            
                            −
                            n
                            +
                            1
                          
                          )
                        
                        +
                        ⋯
                        +
                        
                          (
                          
                            
                              n
                              
                                2
                              
                            
                            +
                            n
                            −
                            1
                          
                          )
                        
                      
                      ⏟
                    
                  
                  
                    
                      n
                      
                        3
                      
                    
                  
                
              
            
            
              
              
                
                =
                
                  
                    
                      
                        
                          
                            
                              
                                
                                  
                                    
                                      
                                        
                                          
                                            
                                              1
                                              ⏟
                                            
                                          
                                          
                                            
                                              1
                                              
                                                2
                                              
                                            
                                          
                                        
                                        +
                                        3
                                      
                                      ⏟
                                    
                                  
                                  
                                    
                                      2
                                      
                                        2
                                      
                                    
                                  
                                
                                +
                                5
                              
                              ⏟
                            
                          
                          
                            
                              3
                              
                                2
                              
                            
                          
                        
                        +
                        ⋯
                        +
                        
                          (
                          
                            
                              n
                              
                                2
                              
                            
                            +
                            n
                            −
                            1
                          
                          )
                        
                      
                      ⏟
                    
                  
                  
                    
                      
                        (
                        
                          
                            
                              
                                n
                                
                                  2
                                
                              
                              +
                              n
                            
                            2
                          
                        
                        )
                      
                      
                        2
                      
                    
                  
                
              
            
            
              
              
                
                =
                (
                1
                +
                2
                +
                ⋯
                +
                n
                
                  )
                  
                    2
                  
                
              
            
            
              
              
                
                =
                
                  
                    (
                  
                
                
                  ∑
                  
                    k
                    =
                    1
                  
                  
                    n
                  
                
                k
                
                  
                    
                      )
                    
                  
                  
                    2
                  
                
                .
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}\sum _{k=1}^{n}k^{3}&=1+8+27+64+\cdots +n^{3}\\&=\underbrace {1} _{1^{3}}+\underbrace {3+5} _{2^{3}}+\underbrace {7+9+11} _{3^{3}}+\underbrace {13+15+17+19} _{4^{3}}+\cdots +\underbrace {\left(n^{2}-n+1\right)+\cdots +\left(n^{2}+n-1\right)} _{n^{3}}\\&=\underbrace {\underbrace {\underbrace {\underbrace {1} _{1^{2}}+3} _{2^{2}}+5} _{3^{2}}+\cdots +\left(n^{2}+n-1\right)} _{\left({\frac {n^{2}+n}{2}}\right)^{2}}\\&=(1+2+\cdots +n)^{2}\\&={\bigg (}\sum _{k=1}^{n}k{\bigg )}^{2}.\end{aligned}}}
  


= Tong cua cac lap phuong le đau tien =
Tong cua n lap phuong le đau tien la so tam giac thu 2n2 − 1:

  
    
      
        
          ∑
          
            k
            =
            1
          
          
            n
          
        
        (
        2
        k
        −
        1
        
          )
          
            3
          
        
        =
        2
        
          n
          
            4
          
        
        −
        
          n
          
            2
          
        
        =
        
          C
          
            2
            
              n
              
                2
              
            
          
          
            2
          
        
      
    
    {\displaystyle \sum _{k=1}^{n}(2k-1)^{3}=2n^{4}-n^{2}=C_{2n^{2}}^{2}}
  

Trong đo, 
  
    
      
        
          C
          
            2
            
              n
              
                2
              
            
          
          
            2
          
        
      
    
    {\displaystyle C_{2n^{2}}^{2}}
  
 la to hop chap 2 cua 2n2.


Trong ly thuyet so = Bai toan Waring đoi voi so lap phuong =

Moi so nguyen co the viet thanh tong cua chin (hoac it hon) so lap phuong nguyen duong. Gia tri chan tren khong the giam đi đuoc boi, vi du nhu 23 khong the viet thanh tong cua it hon chin so lap phuong:

23 = 23 + 23 + 13 + 13 + 13 + 13 + 13 + 13 + 13.


= Tong cua ba so lap phuong =

Hien tai đang co gia thuyet so nguyen khong đong du bang ±4 voi 9 co the viet thanh tong cua ba so lap phuong trong vo han cach. Vi du, 
  
    
      
        6
        =
        
          2
          
            3
          
        
        +
        (
        −
        1
        
          )
          
            3
          
        
        +
        (
        −
        1
        
          )
          
            3
          
        
      
    
    {\displaystyle 6=2^{3}+(-1)^{3}+(-1)^{3}}
  
. Cac so nguyen đong du voi ±4 modulo 9 khong can xet vi chung khong the viet thanh tong cua ba so lap phuong.
So nguyen duong nho nhat ma chua tim đuoc tong la 114. Vao thang chin nam 2019, so nguyen duong nho nhat đung truoc khong tim đuoc tong, so 42, thoa man phuong trinh:

  
    
      
        42
        =
        (
        −
        80538738812075974
        
          )
          
            3
          
        
        +
        
          80435758145817515
          
            3
          
        
        +
        
          12602123297335631
          
            3
          
        
        .
      
    
    {\displaystyle 42=(-80538738812075974)^{3}+80435758145817515^{3}+12602123297335631^{3}.}
  


= Đinh ly cuoi cung cua Fermat đoi voi lap phuong =
Phuong trinh x3 + y3 = z3 khong co nghiem nguyen khac khong (tuc xyz = 0). Tham chi, no con khong co nghiem dang so nguyen Eisenstein.
Ca hai y tren cung đung voi phuong trinh x3 + y3 = 3z3.


So thuc, so phuc Cho ham x ↦ x3: R → R. Chi co ba so bang lap phuong cua chinh minh: -1, 0, va 1. Neu -1 <x <0 hoac 1 <x, thi x3> x. Neu x <-1 hoac 0 <x <1, thi x3 <x. Tinh chat noi tren cung đung voi bat ky so mu le cao hon (x5, x7,...) cua so thuc.
Voi nhung so phuc, lap phuong cua mot so thuan ao la: i3 = −i.


Lich su Cac nha toan hoc Luong Ha đa tao ra cac vien nen hinh nem voi cac ban đe tinh cac khoi lap phuong va cac khoi lap phuong theo thoi ky Babylon (the ky XX đen XVI TCN). Phuong trinh bac ba đuoc nha toan hoc nguoi Hy Lap co la Diophantus biet đen. Anh hung cua Alexandria đa nghi ra mot phuong phap tinh toan coi nguon cua lap phuong vao the ky đau tien cua Cong Nguyen. Phuong phap giai phuong trinh bac ba va phep khai can bac ba xuat hien trong cuu chuong toan thuat, cong trinh toan hoc Trung Quoc đuoc bien soan vao khoang the ky thu II truoc cong nguyen, đuoc Luu Huy chu giai vao the ky thu III cua Cong nguyen. Nha toan hoc nguoi An Đo, Aryabhata đa viet mot loi giai thich ve lap phuong trong nghien cuu cua ong. Trong nam 2010 Alberto Zanoni đa tim ra mot thuat toan moi đe tinh toan lap phuong cua mot so nguyen dai trong mot pham vi nhat đinh, nhanh hon gap đoi.


Tham khao