How do you factor #27m^3+1#?

2 Answers
Jul 3, 2016

Answer:

#(3m+1)(9m^2-3m+1)#

Explanation:

This is a #color(blue)"sum of cubes"# and is factorised in general as follows.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^3+b^3=(a+b)(a^2-ab+b^2))color(white)(a/a)|)))#

#27m^3=(3m)^3" and " 1=(1)^3rArra=3m" and " b=1#

Substituting these values gives required factors.

#rArr27m^3+1=(3m+1)(9m^2-3m+1)#

Jul 3, 2016

Answer:

Using the formula for the sum of cubes.

Explanation:

We can use the formula for the sum of cubes

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

here our cubes are

#(3m)^3+1^3=(3m+1)((3m)^2-3m+1)#

#=(3m+1)(9m^2-3m+1)#.

The roots of #9m^2-3m+1# are

#m=(3\pm sqrt(9-36))/(18)=(3\pm sqrt(-27))/18#

#=(3\pm i3sqrt(3))/18=(1\pm isqrt(3))/6#

then are both complex. This means that the quadratic part cannot be factorized more with real numbers. We can write the irreducible factorization in complex numbers as

#(3m+1)(m-(1+isqrt(3))/6)(m-(1-isqrt(3))/6)#.