How do you factor #y= 3x^3 + 81 # ?

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Answer 1 · 2018-03-06T13:36:53

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

#y=3x^3+81#

#=>y=3(x^3+27)#

#=>y=3(x^3+3^3)#

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

So , #y=3(x^3+3^3)#

#=> y=3(x+3)(x^2+3^2-x xx 3)#

#=> y=3(x+3)(x^2+9-3x)#