Short cut cubing of any three digit number just in few seconds with help of Horner`s method for polynomial evaluation

We can utilize following algebraic expansion of Horner`s method for polynomial evaluation to easily find cube of any three digit number

(z + d) ³ = z x [z x (z + 3d) + 3d²] + d³

Where, d will always be one of the numbers between (+/-) 1 to (+/-) 50.

For example,

To find cube of the number 323,

Let, z = 300 and d = 23

So,

(z + d) ³ = z x [z x (z + 3d) + 3d²] + d³

(300 + (23)) ³ = 300 x [300 x (300 + 3(23)) + 3(23)²] + (23)³

(323)³ = 300 x [300 x (300 + 69) + 3(23)²] + (23)³

= 300 x [300 x 369 + 3(23)²] + (23)³

= 300 x [300 x 369 + 1587] + (23)³

= 300 x [110700 + 1587] + (23)³

= 300 x 112287 + (23)³

= 33686100 + 12167

= 33698267

Short cut cubing of any two digit number just in few seconds with help of Horner`s method for polynomial evaluation

We can utilize following algebraic expansion of Horner`s method for polynomial evaluation to easily find cube of any two digit number

(z + d) ³ = z x [z x (z + 3d) + 3d²] + d³

Where, d will always be one of the numbers (+/-)1, (+/-)2, (+/-)3, (+/-)4, (+/-)5.

So, 3d2 will always be one of the numbers 3, 12, 27, 48, or 75.

For example,

To find cube of the number 23,

Let, z = 20 and d = 3

So,

(z + d) ³ = z x [z x (z + 3d) + 3d²] + d³

(20 + 3)³ = 20 x [20 x (20 + 3(3)) + 3(3)²] + (3)³

(23)³ = 20 x [20 x (20 + 9) + 27] + (3)³

= 20 x [20 x 29 + 27] + (3)³

= 20 x [580 + 27] + (3)³

= 20 x 607 + (3)³

= 12140 + 27

= 12167

Another example,

To find cube of the number 88,

Let, z = 90 and d = (-2)

So,

(z + d) ³ = z x [z x (z + 3d) + 3d²] + d³

(90 + (-2)) ³ = 90 x [90 x (90 + 3(-2)) + 3(-2)²] + (-2)³

(88)³ = 90 x [90 x (90 + (-6)) + 12] + (-2)³

= 90 x [90 x 84 + 12] + (-2)³

= 90 x [7560 + 12] + (-2)³

= 90 x 7572 + (-2)³

= 681480 + (-8)

= 681472

Short cut cubing of any two digit number just in few seconds with help of algebraic identity

We can utilize following algebraic identity to easily find cube of any two digit number

A³ = (A - d) x A x (A + d) + d² x A

Where, d = any assumed value to easily compute cube.

Naturally, this formula works for any value of d, but we choose d to be the distance to a number close to A that is easy to multiply.
For example,

To find cube of the number 23,

Let, d = 3, A = 23

So,

A³ = (A - d) x A x (A + d) + d² x A

= (23 - 3) x 23 x (23 + 3) + (3)² x 23

= 20 x 23 x 26 + (3)² x 23

= 20 x 598 + 9 x 23

= 11960 + 207

= 12167

Short cut squaring of any three digit number just in few seconds with help of algebraic identity

We can utilize following algebraic identity to easily find square of any three digit number

A² = (A - d) x (A + d) + d²

Where, d = any assumed value to easily compute square.

We can quickly square three-digit numbers by rounding up and down to the nearest hundred. 
For example,

To find square of the number 223,

Let, d = 23, A = 223

So,

A² = (A - d) x (A + d) + d²

= (223 - 23) x (223 + 23) + (23)²

= 200 x 246 + 529

= 49200 + 529

             = 49729

Short cut squaring of any two digit number just in few seconds with help of algebraic identity

We can utilize following algebraic identity to easily find square of any two digit number

A² = (A - d) x (A + d) + d²

Where, d = any assumed value to easily compute square.

Naturally, this formula works for any value of d, but we choose d to be the distance to a number close to A that is easy to multiply. 
For example,

To find square of the number 23,

Let, d = 3, A = 23

So,

A² = (A - d) x (A + d) + d²

= (23 - 3) x (23 + 3) + (3)²

= 20 x 26 + 9

= 520 + 9

= 529

Another example,

To find square of the number 48,

Let, d = 2, A = 48

So,

A2 = (A - d) x (A + d) + d²

= (48 - 2) x (48 + 2) + (2)²

= 46 x 50 + 4

= 2300 + 4

= 2304