Ta có: \(\left(a-1\right)^3=9a\)
=>\(a^3-3a^2+3a-1=9a\)
=>\(a^3-1=3a^2-3a+9a=3a^2+6a\)
=>\(\left(a-1\right)\left(a^2+a+1\right)=3a+3\left(a^2+a\right)\)
=>(a-1)(b+1)=3a+3b=3(a+b)
=>\(\left(a-1\right)^3\cdot\left(b+1\right)^3=\left\lbrack3\left(a+b\right)\right\rbrack^3=27\cdot\left(a+b\right)^3\)
Ta có: \(a^3-3a^2+3a-1=9a\)
=>1+3(a+1)+\(3\left(a+1\right)^2+\left(a+1\right)^3=9\left(a+1\right)^2\)
=>\(a^3+3a^3\left(a+1\right)+3a^3\left(a+1\right)^2+a^3\cdot\left(a+1\right)^3=a^3\cdot9\left(a+1\right)^2\)
=>\(a^3+3a^2\cdot\left\lbrack a\left(a+1\right)\right\rbrack+3\cdot a\cdot\left\lbrack a\left(a+1\right)\right\rbrack^2+\left\lbrack a\left(a+1\right)\right\rbrack^3=a^2\cdot\left(a+1\right)^2\cdot9a\)
=>\(a^3+3a^2b+3ab^2+b^3=9ab^2\)
=>\(\left(a+b\right)^3=9ab^2=\left(a-1\right)^3\cdot b^2\)
=>\(27\left(a+b\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(a-1\right)^3\cdot\left(b+1\right)^3=27\left(a-1\right)^3\cdot b^2\)
=>\(\left(b+1\right)^3=27b^2\)
