c) Ta có:
\(VT=\left(a+b+c\right)^2+\left(a-b+c\right)^2+\left(a+b-c\right)^2+\left(-a+b+c\right)^2\)
\(=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+\left(a^2+b^2+c^2-2ab-2bc+2ca\right)+\left(a^2+b^2+c^2+2ab-2bc-2ca\right)+\left(a^2+b^2+c^2-2ab+2bc-2ca\right)\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2-2ab-2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ca+a^2+b^2+c^2-2ab+2bc-2ca\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)=VP\) (đpcm)
d) Ta có:
\(VT=\left(x+y\right)^4+x^4+y^4\)
\(=\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)+x^4+y^4\)
\(=x^4+4x^3y+6x^2y^2+4xy^3+y^4+x^4+y^4\)
\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)
\(VP=2\left(x^2+xy+y^2\right)^2\)
\(=2\left(x^4+x^2y^2+y^4+2x^3y+2xy^3+2x^2y^2\right)\)
\(=2x^4+2x^2y^2+2y^4+4x^3y+4xy^3+4x^2y^2\)
\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)
Từ đó, suy ra: \(VT=VP\) (đpcm)
e) Ta có:
\(VT=\left(5a-3b+8c\right)\left(5a-3b-8c\right)\)
\(=\left[\left(5a-3b\right)+8c\right]\left[\left(5a-3b\right)-8c\right]\)
\(=\left(5a-3b\right)^2-64c^2\)
\(=\left(5a-3b\right)^2-16.4c^2\)
\(=\left(5a-3b\right)^2-16\left(a^2-b^2\right)\) (vì \(a^2-b^2=4c^2\))
\(=25a^2-30ab+9b^2-16a^2+16b^2\)
\(=9a^2-30ab+25b^2\)
\(=\left(3a-5b\right)^2=VP\) (đpcm)
f) Theo giả thiết, ta có:
\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
\(\Rightarrow a=b=c\) (đpcm)
