Learn and know what is the formula of * a minus b whole cube* in algebra chapter. Do you know this formula we can give as an example of identity in math.

Get the * a minus b whole cube formula* by doing a small modification i.e. just replacing the “plus” sign to the “minus” sign in the

*a plus b whole cube formula*. Otherwise in the normal method also we can derive the

**formula. So first we will know what is the { \left( a-b \right) }^{ 3 } formula and later we will see the derivation of it.**

*a minus b whole cube***Formula for a minus b whole cube as follows:**

* a minus b whole cube* is written as

*a*cube minus 3

*a*square

*b*plus 3

*a b*square minus

*b*cube. Mathematically we can express the formula as { \left( a-b \right) }^{ 3 } = ( { a }^{ 3 } – 3 × { a }^{ 2 } ×

*b*+ 3 ×

*a*× { b }^{ 2 } – { b }^{ 3 } )

**Derivation of a minus b whole cube formula:**

We can write

{ \left( a-b \right) }^{ 3 }

= { \left( a-b \right) }^{ 2 } × (*a – b*)

= ( { a }^{ 2 } – 2 × *a × b* + { b }^{ 2 } ) × (*a – b*)

= ( { a }^{ 3 } – { a }^{ 2 } × *b* – 2 × { a }^{ 2 } × *b* + 2 ×* a* × { b }^{ 2 } + *a* × { b }^{ 2 } – { b }^{ 3 } )

= ( { a }^{ 3 } – 3 × { a }^{ 2 } × *b* + 3 × *a* × { b }^{ 2 } – { b }^{ 3 } ) (adding like terms)

Therefore, { \left( a-b \right) }^{ 3 } ** = ( { a }^{ 3 } – 3 × { a }^{ 2 } × **

*b*+ 3 ×

*a*

**× { b }^{ 2 } – { b }^{ 3 } )**

On simplification, we can also write the formula as { \left( a-b \right) }^{ 3 } ** = { a }^{ 3 } – { b }^{ 3 } – ***3 × a × b × (a-b)*

**Examples:**

Solve: { \left( 2x-4 \right) }^{ 3 }

Solution:

We know that,

{ \left( a-b \right) }^{ 3 } = ( { a }^{ 3 } – 3 × { a }^{ 2 } ^{ }× *b* + 3 × *a* × { b }^{ 2 } – { b }^{ 3 } )

{ \left( 2x-4 \right) }^{ 3 } = { \left( 2x \right) }^{ 3 } – 3 { \left( 2x \right) }^{ 2 } 4 + 3 (2*x*) { \left( 4 \right) }^{ 2 } – { 4 }^{ 3 }

{ \left( 2x-4 \right) }^{ 3 } = 8 { x }^{ 3 } – 48 { x }^{ 2 } + 96*x* – 64

Solve: { \left( x-6 \right) }^{ 3 }

Solution:

We know that,

{ \left( a-b \right) }^{ 3 } = ( { a }^{ 3 } – 3 × { a }^{ 2 } ^{ }× *b* + 3 × *a* × { b }^{ 2 } – { b }^{ 3 } )

{ \left( x-6 \right) }^{ 3 } = { \left( x \right) }^{ 3 } – 3 { \left( x \right) }^{ 2 } 6 + 3 (*x*) { \left( 6 \right) }^{ 2 } – { 6 }^{ 3 }

{ \left( x-6 \right) }^{ 3 } = { \left( x \right) }^{ 3 } – 18 { x }^{ 2 } + 108*x* -216

Solve: { \left( 5y-3x \right) }^{ 3 }

Solution:

We know that,

{ \left( a-b \right) }^{ 3 } = ( { a }^{ 3 } – 3 × { a }^{ 2 } ^{ }× *b* + 3 × *a* × { b }^{ 2 } – { b }^{ 3 } )

{ \left( 5y-3x \right) }^{ 3 } = { \left( 5y \right) }^{ 3 } – 3 { \left( 5y \right) }^{ 2 } 3*x* + 3 (5y) { \left( 3x \right) }^{ 2 } – { \left( 3x \right) }^{ 3 }

= 125 { y }^{ 3 } – 225*x* { y }^{ 2 } + 135 { x }^{ 2 } y – 27 { x }^{ 3 }

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