How do you factor #8(5x + 7)^3 + 27(x-9)^3#?

1 Answer
George C.
Dec 31, 2015

Use the sum of cubes identity to find:

#8(5x+7)^3 + 27(x-9)^3 = 13(x-1)(79x^2+346x+1303)#

Explanation:

The sum of cubes identity may be written:

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

Use this with #a=10x+14# and #b=3x-27# as follows...

#8(5x+7)^3 + 27(x-9)^3#

#= 2^3(5x+7)^3 + 3^3(x-9)^3#

#= (10x+14)^3+(3x-27)^3#

#= ((10x+14)+(3x-27))((10x+14)^2-(10x+14)(3x-27)+(3x-27)^2)#

#= (13x-13)((100x^2+280x+196)-(30x^2-228x-378)+(9x^2-162x+729))#

#=13(x-1)(79x^2+346x+1303)#

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