Question

factorise (a-b)^3+(b-c)^3+(c-a)^3-3(a+b)(b+c)(c+a)

Answer 1

Answer:

Step-by-step explanation:

Input:

(a - b)^3 + (b - c)^3 + (c - a)^3 - 3 (a + b) (b + c) (c + a)

Solutions:

a = -(6 c (b + c) ± sqrt(36 c^2 (b + c)^2 - 144 b^3 c))/(12 b) (b!=0)

Geometric figure:

line

Alternate forms:

-6 (a^2 b + c (a (b + c) + b^2))

-(3 (2 a b + b c + c^2)^2)/(2 b) - (3 (4 b^3 c - b^2 c^2 - 2 b c^3 - c^4))/(2 b)

a (-6 a b - 6 b c - 6 c^2) - 6 b^2 c

Expanded form:

-6 a^2 b - 6 a b c - 6 a c^2 - 6 b^2 c

Polynomial discriminant:

Δ_a = -36 (4 b^3 c - b^2 c^2 - 2 b c^3 - c^4)

Property as a function:

Parity

odd

Derivative:

d/da((a - b)^3 + (b - c)^3 + (c - a)^3 - 3 (a + b) (b + c) (c + a)) = -6 (2 a b + c (b + c))

Indefinite integral:

integral((a - b)^3 + (b - c)^3 + (-a + c)^3 - 3 (a + b) (a + c) (b + c)) da = -6 ((a^3 b)/3 + 1/2 a^2 c (b + c) + a b^2 c) + constant

Definite integral over a cube of edge length 2 L:

integral_(-L)^L integral_(-L)^L integral_(-L)^L ((a - b)^3 + (b - c)^3 + (-a + c)^3 - 3 (a + b) (a + c) (b + c)) dc db da = 0