How do you factor #54x^3 - 2y^3#?

1 Answer
Mar 30, 2018

Answer:

#2(3x-y)(9x^2+3xy+y^2)#

Explanation:

When factoring,
Step 1: Always factor out a GCF first, if possible.

In this particular problem, we can factor out 2:
#54x^3-2y^3# = #2(27x^3-y^3)#

Step 2: Count the number of terms:

—if 2 terms, check if problem is:

difference of squares #(a^2-b^2)#, which factors into #(a-b)(a+b)#

difference of cubes #(a^3-b^3)#, which factors into #(a-b)(a^2+ab+b^2)#

sum of cubes #(a^3+b^3)#, which factors into #(a+b)(a^2-ab+b^2)#

(#27x^3-y^3)# is a difference of cubes:
#27x^3# = #(3x)^3#
#y^3# = #(y)^3#

Plugging these values into the difference of cubes formula:
#(a^3-b^3)# = #(a-b)(a^2+ab+b^2)#
#((3x)^3-((y)^3)# = #((3x)-(y)) ((3x)^2+(3x)(y)+ (y)^2)#
#27x^3-y^3# = #(3x-y)(9x^2+3xy+y^2)#

Putting it all together:

#54x^3-2y^3# = #2(27x^3-y^3)#
#54x^3-2y^3# = #2(3x-y)(9x^2+3xy+y^2)#