Ta có:\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)với\(\forall a,b,c\in R\)
Phương trình: \(x^3-y^3-xy=25\Leftrightarrow\left(3x\right)^3-\left(3y\right)^3-1^3-3.3x.3y.1=25.27-1\)
\(\Leftrightarrow\left(3x-3y-1\right)\left(9x^2+9y^2+1+18xy+6x-6y\right)=674=2.337=337.2\)
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