Factor using the Binomial Theorem? #8a^3+12a^2b+6ab^2+b^3#

Question

2686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-27T20:52:53

#8a^3+12a^2b+6ab^2+b^3 = (2a+b)^3#

Note that #8a^3 = (2a)^3# and #b^3# are both perfect cubes and this polynomial is homogeneous of degree #3#.

So let's expand #(2a+b)^3# and see if it is equal.

#(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k#

where #((n),(k)) = (n!)/((n-k)!k!)#

In particular:

#(x+y)^3 = ((3),(0))x^3+((3),(1))x^2y+((3),(2))xy^2+((3),(3))y^3#

We find:

#((3),(0)) = ((3),(3)) = (3!)/((3!)(0!)) = 1#

#((3),(1)) = ((3),(2)) = (3!)/((2!)(1!)) = 6/2 = 3#

So:

#(x+y)^3 = x^3+3x^2y+3xy^2+y^3#

Then putting #x=2a# and #y=b#, we find:

#(2a+b)^3 = (2a)^3+3(2a)^2(b)+2(2a)(b)^2+(b)^3#

#color(white)((2a+b)^3) = 8a^3+12a^2b+6ab^2+b^3#