Question

(c) Factorise: a3 + b3 + c3 - 3abc.​

Answer 1

a3 + b3 + c3 - 3abc = (a + b +

c)(a² + b² + c² - ab - bc - ca)

Identity: a3 + b3 + c3 3abc = (a + b + c)(a2 +

b2 + c2 - ab - bc - ca)

How is this identity obtained?

Let's see how.

Taking RHS of the identity:

(a + b + c)(a + b2 + c2-ab-bc-ca)

Multiply each term of first polynomial with

every term of second polynomial, as shown

below:

= a(a + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 -

ab-bc - ca ) + c(a + b2 + c2 - ab-bc-ca)

= { (a* a?) + (a*b2) + (a * (?) - (a * ab) - (a*

bc) (a * ca) } + {(b * a) + (b*b2) + (b* c2) - (b

* ab) - (b* bc) - (b * ca)} + {(c* a?) + (b* b2) +

(c * c2) - (c * ab) - (c* bc) - (c* ca)}

Now, solve multiplication in curly braces and

we get:

= a3 + ab + ac2 - a²b - abc a?c + a?b + b3 +

bc2 - ab? - b?c abc + a?c + b?c + c3 - abc -

bc2 - ac?

Rearrange the terms and we get:

= a3 + b3 + c3 + alb - alb + ac2-ac2 + ab2 - ab2

+ bc2 - bc2 + a?c - a?c + b2c - b?c - abc - abc -

abc

Above highlighted like terms will be subtracted

and we get:

= a3 + b3 + c3 - abc - abc - abc

Adding like terms i.e (-abc) and we get:

= a3 + b3 + c3 - 3abc

Hence, a3 + b3 + c3 - 3abc = (a + b + c)(a + b2

+ c2 - ab-bc-ca)

Following are a few applications to this

identity.