How do you factor #c^ { 3} + 64#?

2 Answers
Dec 17, 2016

Answer:

#(c + 4)xx(c^2- 4c + 16)#

Explanation:

There is a Theory:

A sum of two perfect cubes, #a^3+b^3# can be factored into
#(a+b)xx(a^2-ab+b^2)#

So #c^3+64=c^3+4^3#
#(c+4)xx(c^2-4c+4^2)#
#(c+4)xx(c^2-4c+16)#

Dec 17, 2016

Answer:

#(c^3+64) = (c+4)(c^2-4c+16)#

Explanation:

The generalized factoring of #x^3+a^3# is #(x+a)(x^2-ax+a^2)#
as demonstrated below:

#{: (xx," | ",x^2,-ax,a^2), ("-----",,"-----","-----","-----"), (x," | ",x^3,-ax^2,a^2x), (a," | ",ax^2,-a^2x,a^3), ("-----","-----","-----","-----","-----"), (,x^3,,,a^3) :}#