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a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= a(b3 - c3 ) + b( c3 - b3 + b3 - a3) + c(a3 - b3)
= a(b3 - c3) + b(c3 - b3) + b(b3 - a3) + c(a3 - b3)
= a(b3 - c3) - b(b3 - c3) - [b(a3 - b3) - c(a3- b3)]
= (b3 - c3)(a - b) - (a3- b3)(b - c)
= (b - c)(b2 + bc + c2)(a - b) - (a - b)(a2 + ab + b2)(b - c)
= (b - c)(a - b)(b2 + bc + c2 - a2 + ab - b2)
= (b - c)(a - b) [ (c2 - a2) + (bc - ab) ]
= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]
= (b - c)(a -b) [ (c - a)(c + a + b) ]
= (a- b)(b - c)(c - a)(a + b + c)
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a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
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a(b3 - c3) + b(c3 - a3) + c(a3 - b3)
= a(b3 - c3 ) + b( c3 - b3 + b3 - a3) + c(a3 - b3)
= a(b3 - c3) + b(c3 - b3) + b(b3 - a3) + c(a3 - b3)
\(=\left[a\left(b^3-c^3\right)-b\left(b^3-c^3\right)\right]-\left[b\left(a^3-b^3\right)-c\left(a^3-b^3\right)\right]\)
= (b3 - c3)(a - b) - (a3- b3)(b - c)
= (b - c)(b2 + bc + c2)(a - b) - (a - b)(a2 + ab + b2)(b - c)
= (b - c)(a - b)(b2 + bc + c2 - a2 + ab - b2)
= (b - c)(a - b) [ (c2 - a2) + (bc - ab) ]
= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]
= (b - c)(a -b) [ (c - a)(c + a + b) ]
= (a- b)(b - c)(c - a)(a + b + c)
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Theo bài ra, ta có: a+b+c
Suy ra: 3(a+b+c)-3abc=0
Suy ra: -3abc=0
Tương đương: -3*(b+c)*(a+c)*(a+b)=0
Tương đương: -3* a^2+b^2+c^2=0
Tương đương: -3*0=0
Suy ra: nếu a+b+c=0 thì a3+b3+c3-3abc=0(đpcm)
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Chúc bạn học tốt nha!!
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3ab\)
\(=\left[\left(a+b\right)+c\right]\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)