Ta có:
\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2=\left(a+b\right)^3-3ab\left(a+b\right)=3ab-1\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3ab+1=0\)
\(\Leftrightarrow\left(a+b\right)^3+1-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left[\left(a+b\right)^2-\left(a+b\right)+1\right]-3ab\left(a+b+1\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+1\right)\left(a^2+b^2-ab-a-b+1\right)=0\)
=> a+b+1 = 0 hoặc a^2+b^2-ab-a-b-1=0
<=> a+b = -1 hoặc (a-b)^2+b(a-1)-(a+1) = 0
<=> a+b = -1 hoặc (a-b)^2-(a+1)(b-1)=0
....
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