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ab(a+b) + bc(b+c) + ca(c+a) = a^2b + ab^2 + b^2c + bc^2 + ca(c+a) + 2abc
= ab^2 + b^2c + a^2b + bc^2 + 2abc + ca(c+a)
=b^2(a+c) + b(a^2 + c^2 + 2ac) + ca(c+a)
=b^2(a+c) + b(a+c)^2 + ca(c+a)
=(c+a)[b^2 + b(a+c) + ca]
=(c+a)[b^2 + ab + bc + ca]
=(c+a)[b(b+a) + c(b+a)]
=(c+a)(b+c)(b+a)
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.\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)
=\(a\left(b^2-2bc+c^2-a^2\right)+b\left(a^2+2ac+c^2-b^2\right)+c\left(a^2-2ab+b^2-c^2\right)\)
=\(a\left[\left(b-c\right)^2-a^2\right]+b\left[\left(a+c\right)^2-b^2\right]+=c\left[\left(a-b^2\right)-c^2\right]\)
=\(a\left(c-b+a\right)\left(a+b-c\right)+b\left(a+c-b\right)\left(a+b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
=\(\left(a+c-b\right)\left[a\left(c-b+a\right)+b\left(a+b+c\right)+c\left(a-b-c\right)\right]\)
=\(\left(a+c-b\right)\left(b+a-c\right)\left(c+b-a\right)\)
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Ta có:\(a\cdot\left(b^2+c^2\right)+b\cdot\left(c^2+a^2\right)+c\cdot\left(a^2+b^2\right)+2abc\\\)
=>\(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2+2abc\)
=>\(\left(a^2b+ab^2+a^2c+abc\right)+\left(abc+c^2a+cb^2+c^2b\right)\)
Đặt a ra ngoài ở số hạng thứ nhất và c ra ngoài ở số hạng thứ hai ta được:
\(a\cdot\left(ab+b^2+ac+bc\right)+c\cdot\left(ab+ac+b^2+bc\right)\)
=>\(a\cdot\left(b+c\right)\cdot\left(a+b\right)+c\cdot\left(b+c\right)\cdot\left(a+b\right)\)
=>\(\left(a+c\right)\cdot\left(a+b\right)\cdot\left(b+c\right).\)
phân tích đa thức thành nhân tử
\(ab.\left(a+b\right)+bc.\left(b+c\right)+ac.\left(c+a\right)+2abc\)
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\(ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)+2abc\)
\(=ab\left(a+b\right)+b^2c+bc^2+a^2c+ac^2+2abc\)
\(=ab\left(a+b\right)+\left(ac^2+bc^2\right)+\left(a^2c+2abc+b^2c\right)\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a^2+2ab+b^2\right)\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+\left(ac+bc\right)\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=\left(a+b\right)\left[\left(ab+ac\right)+\left(c^2+bc\right)\right]\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
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\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=\)\(ab\left(a+b\right)+bc\left(a+b+c-a\right)+ca\left(c+a\right)+2abc\)
\(=\)\(ab\left(a+b\right)+bc\left(a+b\right)+bc\left(c-a\right)+ca\left(c+a\right)+2abc\)
\(=\)\(\left(a+b\right)\left(ab+bc\right)+bc\left(c-a\right)+ca\left(c+a\right)+2abc\)
\(=\)\(b\left(a+b\right)\left(c+a\right)+bc\left(c-a\right)+ca\left(c+a\right)+2abc\)
\(=\)\(b\left(ac+a^2+bc+ab\right)+b\left(c^2-ca\right)+ca\left(c+a\right)+2abc\)
\(=\)\(b\left(ca+a^2+bc+ab+c^2-ca\right)+ca\left(c+a\right)+2abc\)
\(=\)\(b\left(a^2+ab+bc+c^2\right)+ca\left(c+a\right)+2abc\)
\(=\)\(b\left(a^2+2ca+c^2+ab+bc\right)+ca\left(c+a\right)\)
\(=\)\(b\left[\left(c+a\right)^2+b\left(c+a\right)\right]+ca\left(c+a\right)\)
\(=\)\(b\left(c+a\right)\left(a+b+c\right)+ca\left(c+a\right)\)
\(=\)\(\left(c+a\right)\left(ab+b^2+bc+ca\right)\)
\(=\)\(\left(c+a\right)\left[b\left(a+b\right)+c\left(a+b\right)\right]\)
\(=\)\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
...
cách này ngắn hơn nè:
\(ab.\left(a+b\right)+bc.\left(b+c\right)+ac.\left(a+c\right)+2abc\)
\(=a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+abc+abc\)
\(=\left(abc+ac^2\right)+\left(abc+b^2c\right)+\left(a^2b+ab^2\right)+\left(c^2a+c^2b\right)\)
\(=ac.\left(a+b\right)+bc.\left(a+b\right)+ab.\left(a+b\right)+c^2.\left(a+b\right)\)
\(=\left(a+b\right).\left(ac+bc+ab+c^2\right)\)
\(=\left(a+b\right).\left[c\left(a+c\right)+b.\left(a+c\right)\right]=\left(a+b\right).\left(c+b\right).\left(a+c\right)\)
ab2+ac2+bc2+a2b+a2c+b2c+2abc
=(a2b+ab2+abc)+(a2c+ac2+abc)+(b2c+bc2)
=ab(a+b+c)+ac(a+b+c)+bc(b+c)
=(a+b+c)(ab+ac)+bc(b+c)
=(a+b+c)a(b+c)+bc(b+c)
=(a+b+c)(b+c)(a+bc)