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\(a\left(b^2-c^2\right)-b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(c^2-a^2\right)-c\left[\left(b^2-c^2\right)+\left(c^2-a^2\right)\right]\)
\(=a\left(b^2-c^2\right)-b\left(c^2-a^2\right)-c\left(b^2-c^2\right)-c\left(c^2-a^2\right)\)
\(=\left(b^2-c^2\right)\left(a-c\right)-\left(c^2-a^2\right)\left(b +c\right)\)
\(=\left(b+c\right)\left(b-c\right)\left(a-c\right)-\left(c-a\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a-c\right)\left(b+c\right)\left(b-c+a+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)
Bn tham khảo lời giải ở câu này nha
https://olm.vn/hoi-dap/question/705592.html
(a2 - b2 -c2 )2 -4a2b2
= (a2 -b2 -c2 )2 - (2ab)2
= (a2 -b2 - c2 -2ab)( a2 -b2-c2 +2ab)
a ) a ³ (b - c) + b ³ (c - a)+ c ³ (a - b)
=a3(b-c)-b3[(b-c)+(a-b)]+c3(a-b)
=a3(b-c)-b3(b-c)-b3(a-b)+c3(a-b)
=(b-c)(a-b)(a2+ab+b2)-(b-c)(a-b)(b2+bc+c2)
=(b-c)(a-b)(a2+2b2+c2+ab+bc)
a) x3+y3+z3-3xyz
=(x+y)3+z3-3x2y-3xy2-3xyz
=(x+y+z).[(x+y)2+(x+y).z+z2]-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2)-3xy.(x+y+z)
=(x+y+z)(x2+2xy+y2+zx+zy+z2-3xy)
=(x+y+z)(x2+y2+zx+zy+z2-zy)
b)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(-a2c+c2a)+(b2c-b2a)
=b.(a2-c2)-ac.(a-c)-b2.(a-c)
=b.(a+c)(a-c)-ac.(a-c)-b2.(a-c)
=(a-c)[b.(a+c)-ac-b2]
=(a-c)(ab+bc-ac-b2)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
ta có :
\(K=a^2\left(b+c\right)+a\left(b^2+c^2+2bc\right)+bc\left(b+c\right)=a^2\left(b+c\right)+a\left(b+c\right)^2+bc\left(b+c\right)\)
\(=\left(b+c\right)\left(a^2+a\left(b+c\right)+bc\right)=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
tương tự L và M có dạng giống hệt K nên ta có
\(\hept{\begin{cases}L=\left(x+y\right)\left(x+z\right)\left(y+z\right)\\M=\left(a+b\right)\left(a+c\right)\left(b+c\right)\end{cases}}\)
= a^3 (b-c) + b^3 ( c- b + b - a) + c^3 ( a-b)
= a^3 (b-c) - b^3 ( b-c) - b^3(a-b) + c^3(a-b)
= (b-c)(a^3 - b^3) - (a-b)(b^3 - c^3)
=(b-c)(a-b)(a^2+ab+b^2) - (a-b)(b-c)(b^2+bc+c^2)
= (a-b)(b-c)(a^2 + ab + b^2 - b^2 - bc - c^2)
= (a-b)(b-c)( a^2 - c^2 + ab - bc)
=(a-b)(b-c)[(a-c)(a+c) + b(a-c)]
=(a-b)(b-c)(a-c)(a+b+c)