Question

a cube+ b cube+ c cube-3ab-1=?

Answer 1

(a+b)3=a3+3a2b+3ab2+b3

So we have

a3+b3−1+3ab=(a+b)3−3a2b−3ab2−1+3ab=(a+b)3−1−3ab(a+b−1)

Using the relationship

(x3−y3)=(x−y)(x2+xy+y2)

(a+b)3−1=(a+b)3−(1)3=(a+b−1)((a+b)2+(a+b)+1)

so we have

(a+b)3−1−3ab(a+b−1)=(a+b−1)((a+b)2+(a+b)+1)−3ab(a+b−1)

or

(a+b-1)\left((a+b)^2+a+b+1-3ab\right) = (a+b-1)\left(a^2+b^2-ab+a+b+1)

so

a^3+b^3-1+3ab=(a+b-1)\left(a^2+b^2-ab+a+b+1)