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a^3(c−b^2)+b^3(a−c^2)+c^3(b−a^2)+abc(abc−1)
=a^3c−a^3b^2+b^3(a−c^2)+bc^3−a^2c^3+a^2b^2c^2−abc
=(a^3c−a^2c^3)+b^3(a−c^2)−(a^3b^2−a^2b^2c^2)+(bc^3−abc)
=a^2c(a−c^2)+b^3(a−c^2)−a^2b^2(a−c^2)−bc(a−c^2)
=(a^2c+b^3−a^2b^2−bc)(a−c2)
=[c(a^2−b)−b^2(a^2−b)](a−c^2)=(a^2-b)(c-b^2)(a-c^2)
1) \(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
a3(c - b2) + b3(a - c2) + c3(b - a2) + abc(abc - 1)
= a3c - a3b2 + ab3 - b3c2 + bc3 - a2c3 + a2b2c2 - abc
= a2b2c2 - b3c2 - (a2c3 - bc3) - (a3b2 - ab3) + (a3c - abc)
= b2c2(a2 - b) - c3(a2 - b) - ab2(a2 - b) + ac(a2 - b)
= (a2 - b)(b2c2 - c3 - ab2 + ac) = (a2 - b)[c2(b2 - c) - a(b2 - c)] = (a2 - b)(b2 - c)(c2 - a)
a) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3b-a^3c+b^3\left(c-a\right)+c^3a-c^3b\)
\(=\left(a^3b-c^3b\right)+\left(c^3a-a^3c\right)+b^3\left(c-a\right)\)
\(=-b\left(c^3-a^3\right)+ca\left(c^2-a^2\right)+b^3\left(c-a\right)\)
\(=-b\left(c-a\right)\left(c^2-ac+a^2\right)+ca\left(c+a\right)\left(c-a\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b\right)+\left(c-a\right)\left(c^2a+ca^2\right)+b^3\left(c-a\right)\)
\(=\left(c-a\right)\left(-c^2b+abc-a^2b+c^2a+ca^2+b^3\right)\)
a) a3 (b-c) + b3 (c-a) +c3 (a-b)
<=> a3b – a3c +b3c – b3a + c3a – c3b
<=> b(a3 – c3) +c(a3 – b3) + a(b3 - c3)
(Tự áp dụng hằng đẳng thức)
b)
Đặt: \(a-b=x;\)\(b-c=y;\)\(c-a=z\)
thì: \(x+y+z=0\)
Dễ dàng chứng minh đc:
\(x+y+z=0\)
thì \(x^3+y^3+z^3=3xyz\)
đến đây bạn thay trở lại nhé
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)
\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)
\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)
\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)
Câu hỏi của Access_123 - Toán lớp 8 - Học toán với OnlineMath