Maharashtra State BoardSSC (English Medium) 8th Standard
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Factorisation using Identity a3 + b3 = (a + b)(a2 - ab + b2)

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Topics

formula

  • a+ b3 = ( a + b )(a2 - ab + b2)

notes

Factorisation using Identity a3 + b3 = (a + b)(a2 - ab + b2):

We know that, (a + b)3 = a3 + 3a2b + 3ab2 + b3, which we can write as 

(a + b)3 = a3 + b3 + 3ab(a + b)

Now, a3 + b3 + 3ab(a + b) = (a + b)3   .....(Interchanging the sides)

∴ a3 + b3 = (a + b)3 - 3ab(a + b)

= [(a + b)(a + b)2] - 3ab(a + b)

= (a + b)[(a + b)2 - 3ab] 

= (a + b)(a2 + 2ab + b2 - 3ab)

= (a + b)(a2 - ab + b2)

a3 + b3 = (a + b)(a2 - ab + b2).

Example

Factorise: x3 + 27y3

x3 + 27y3

= x3 + (3y)3

= (x + 3y)[x2 - x(3y) + (3y)2]

= (x + 3y)[x2 - 3xy + 9y2]

Example

Factorise: 8p3 + 125q3

8p3 + 125q

= (2p)3 + (5q)3

= (2p + 5q)[(2p)2 - 2p × 5q + (5q)2]

= (2p + 5q)(4p2 - 10pq + 25q2)

Example

Factorise: `m^3 + 1/(64m^3)`

`m^3 + 1/(64m^3)`

`= m^3 + (1/(4m))^3`

`= (m + 1/(4m))[m^2 - m xx 1/(4m) + (1/(4m))^2]`

`= (m + 1/(4m))(m^2 - 1/4 + 1/(16m^2))`

Example

Factorise: 250p3 + 432q3

250p3 + 432q3 

= 2(125p3 + 216q3)

= 2[(5p)3 + (6q)3]

= 2(5p + 6q)(25p2 + 30pq + 36q2)

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