How do you factor completely x3-1?
Factorise the given expression using algebraic identity
According to the algebraic identities,
a3-b3=(a-b)(a2+ab+b2)
So, the given expression can be factorised as follows,
x3-1=x3-13
ā‡’x3-1=x-1x2+xĆ—1+12
ā‡’x3-1=x-1x2+x+1
Hence, the factors of x3-1 obtained by using algebraic identity is x-1x2+x+1.
How do you factor completely x3+y3+z3-3xyz?