Factorize b³-3b²+2b-a³+6a²-11a+6?

Question

Dragan, can you show me your method?

DD, I followed you until your sixth line (third from the bottom.) How did you get that?

Answer 1

(b-a+1)*(a^2+b^2+a*b-5*a-4*b+6)

Answer 2

Hi,

b³-3b²+2b-a³+6a²-11a+6 =

b(b - 1)(b - 2) - (a - 1)(a - 2)(a - 3) <==ANSWER

I hope that helps!! :-)

Answer 3

b³-3b²+2b-a³+6a²-11a+6

= b³-3b²+3b -1 -a³+6a²-11a+6 - b + 1

= (b -1)^3 -  b + 1- { a - 3.(a^2).2 + 3.a.(2^2) - 2^3 - a + 8 - 6}

= (b -1)^3 - (b-1) - { ( a - 2)^3 - ( a - 2)}

= { ( b - 1)^ 3 - ( a - 2)^ 3 } - { ( b - 1) - ( a - 2)}

= ( b - a + 1) *{ (b -1)^2 + (b -1)*(a -2) + (a-2)^2  -1}

= (b -a +1)* ( b^ 2 - 2b + 1 + ab -2b - a + 2 + a^2 - 4a + 4 -1}

= ( b -a +1)* (b^2 +a^2 + ab - 4b -5a +4)