\(\left(x+3\right)^3-\left(x+1\right)^3=0\) (1)
\(ĐKXĐ:x\inℝ\)
Pt (1) \(\Leftrightarrow\left(x+3-x-1\right)\left[\left(x+3\right)^2+\left(x+3\right)\left(x+1\right)+\left(x+1\right)^2\right]=0\)
\(\Leftrightarrow2.\left[x^2+6x+9+x^2+4x+3+x^2+2x+1\right]=0\)
\(\Leftrightarrow3x^2+12x+13=0\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+1=0\)
\(\Leftrightarrow3\left(x+2\right)^2+1=0\) ( vô nghiệm do \(3\left(x+2\right)^2+1\ge1>0\forall x\) )
Vậy pt đã cho vô nghiệm.
Áp dụng \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)ta được
\(\left(x+3\right)^3-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left[\left(x+3\right)-\left(x+1\right)\right]^3+3\left(x+3\right)\left(x+1\right)\left[\left(x+3\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow2^3+3\left(x^2+4x+3\right).2=0\)\(\Leftrightarrow8+6\left(x^2+4x+3\right)=0\)
\(\Leftrightarrow6\left(x^2+4x+3\right)=-8\)\(\Leftrightarrow6\left[\left(x^2+4x+4\right)-1\right]=-8\)
\(\Leftrightarrow6\left(x+2\right)^2-6=-8\)\(\Leftrightarrow6\left(x+2\right)^2=-2\)
Vì \(6\left(x+2\right)^2\ge0\forall x\)
mà \(-2< 0\)\(\Rightarrow\)Vô lý
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
