a minus b whole cube formula explained

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.

a minus b whole cube formula explained

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

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

{ \left( 2x-4 \right) }^{ 3 } = 8 { x }^{ 3 } – 48 { x }^{ 2 } + 96x – 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 } + 108x -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 } 3x + 3 (5y) { \left( 3x \right) }^{ 2 } { \left( 3x \right) }^{ 3 }

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

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