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\(c)\)
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(d)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=[\left(a+b\right)c]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)[a\left(b+c\right)+c\left(b+c\right)]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
A= bc(a+d)(b-c) +ac(b+d)(c-a) + ab(c+d)(a-b)
A= bc(ab+ bd -ac -dc ) + ac(bc+cd -ab-ad )+ab(ac+ad-bc-bd)
A=(ab²c + b²cd -abc² -bdc² ) + (abc² + adc² -a²bc -a²cd ) + (a²bc + a²bd - ab²c -ab²d)
A= (ab²c + cb²d -ab²c-ab²d) + (c²ab -abc² -bdc² +adc² ) + ( a²bd +a²bc -a²bc -a²cd)
A= a²(bd-cd) + b²(cd-ad) + c²(ad-bd)
A=a²d(b-c) + b²d(c-a) + c²d(a-b)
A=d(a²b-a²c + b²c-b²a +c²a-c²b)
A=d[b(a²-c²) + c(b²-a²) + a(c² - b²)]
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)
Ta có : \(A=ab(a-b)+bc(b-c)+ca(c-a)\)
\(\Rightarrow A=ab(a-b)-bc(c-b)+ac(c-a)\)
\(\Rightarrow A=ab(a-b)-bc[(c-a)+(a-b)]+ac(c-a)\)
\(\Rightarrow A=ab(a-b)-bc(a-b)-bc(c-a)+ac(c-a)\)
\(\Rightarrow A=(a-b)(ab-bc)+(c-a)(ac-bc)\)
\(\Rightarrow A=b(a-b)(a-c)-(a-c)c(a-b)\)
\(\Rightarrow A=(a-c)(a-b)(b-c)\)
Chúc học tốt trong kì thi tới :>
Ta có ab(a – b) + bc(b – c) + ca(c – a)
= ab(a – b) + bc[b – a + a – c] + ac(c – a)
= ab(a – b) – bc(a – b) + bc(a – c) – ac(a – c)
= (a – b)(ab – bc) + (a – c)(bc – ac)
= b(a – b)(a – c) – c(a – c)(a – b)
= (a – b)(a – c)(b – c)
Đáp án cần chọn là: A