The value of (a^2-b^2)^3(b^2-c^2)+(c^2-a^2)^3 (a-b)^3+(b-c)^3+(c-a)^3 is:

2. 3(a + b) (b + c) (c + a) Solution: (a^2-b^2)^3(b^2-c^2)+(c^2-a^2)^3 (a-b)^3+(b-c)^3+(c-a)^3 = 3(a^2-b^2)(b^2-c^2)(c^2-a^2) 3(a-b)(b-c)(c-a) [Since x^3+y^3+z^3=3xyz, if x+y+z=0] = 3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a) 3(a-b)(b-c(c)-a) =3(a+b)(b+c)(c+a)

The value of (a^2-b^2)^3(b^2-c^2)+(c^2-a^2)^3 (a-b)^3+(b-c)^3+(c-…

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Question

The value of $\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$ is:

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Answer

3(a + b) (b + c) (c + a)
Solution:
$\frac{(\text{a}^2-\text{b}^2)^3(\text{b}^2-\text{c}^2)+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$ $[\text{Since}\text{ x}^3+\text{y}^3+\text{z}^3=3\text{xyz},\text{ if}\text{ x}+\text{y}+\text{z}=0]$
$=\frac{3(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{3(\text{a}-\text{b})(\text{b}-\text{c}(\text{c})-\text{a})}$
$=3(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$