`B=a^3+b^3+6a^2b^2(a+b)+3ab(a^2+b^2)`
`B=(a+b)(a^2+b^2-ab)+6a^2b^2+3ab.(a^2+b^2+2ab-2ab)`
`B=(a^2+b^2+2ab)-3ab+6a^2b^2+3ab.[(a+b)^2-2ab]`
`B=(a+b)^2-3ab+6a^2b^2+3ab(1-2ab)`
`B=1-3ab+6a^2b^2+3ab-6a^2b^2`
`B=1`
Vậy `B=1` khi `a+b=1`
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`a+b=1 <=> a=1-b`
Thay `a=1-b` vào A ta có:
`B=(1-b)^3+b^3+6(1-b)^2.b^2+3(1-b)b[(1-b)^2+b^2]`
`B=1+3b^2-3b-b^3+b^3+6b^2(1-2b+b^2)+(3b-3b^2)(1-2b+b^2+b^2)`
`B=-3b+3b^2+1+6b^2-12b^3+6b^4+3b-6b^2+6b^3-3b^2+6b^3-6b^4`
`B=(6b^4-6b^4)+(-12b^3+6b^3+6b^3)+(3b^2+6b^2-6b^2-3b^2)+(-3b+3b)+1`
`B=1`
Vậy `B=1` khi `a+b=1`
