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\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+xy+yz+zx\right)\)
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Câu hỏi của Trần Điền - Toán lớp 9 - Học toán với OnlineMath
Tham khảo câu b
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b: \(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>(a-c)^2+(a-b)^2+(b-c)^2=0
=>a=b=c
c: \(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)=0\)
=>(a-b)^2+(a-c)^2+(b-c)^2=0
=>a=b=c
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a) ab(a+b) + bc(b+c) + ca(c+a)
= a2b + ab2 + b2c + bc2 + ca(c+a) + 2abc
= ab2 + b2c + a2b + bc2 + 2abc + ca(c+a)
=b2(a+c) + b(a2 + c2 + 2ac) + ca(c+a)
=b2(a+c) + b(a+c)2 + ca(c+a)
=(c+a)[b2 + b(a+c) + ca]
=(c+a)[b2 + ab + bc + ca]
=(c+a)[b(b+a) + c(b+a)]
=(c+a)(b+c)(b+a)
ab2+ac2+bc2+ba2+ca2+b2c+2abc
=bc2+b2c+ac2+abc+ab2+abc+ba2+ca2
=bc(b+c)+ac(b+c)+ab(b+c)+a2(b+c)
=(b+c)(bc+ac+ab+a2)
=(b+c)[c(a+b)+a.(a+b)]
=(b+c)(a+b)(a+c)