simplifying and factoring a fraction

Question

how i get $\frac{(a+b)^2+(a+c)^2+(b+c)^2}{2}$ from $\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-a)(c-b)}$ assuming that $a\ne b\ne c\ne a$

i tried to make $$\begin{align} &\frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-a)(c-b)}\\ &=\frac{a^4}{(a-b)(a-c)}-\frac{b^4}{(a-b)(b-c)}+\frac{c^4}{(a-c)(b-c)}\\ &=\frac{a^4(b-c)}{(a-b)(a-c)(b-c)}-\frac{b^4(a-c)}{(a-b)(a-c)(b-c)}+\frac{c^4(a-b)}{(a-b)(a-c)(b-c)}\\ &=\frac{a^4(b-c)-b^4(a-c)+c^4(a-b)}{(a-b)(a-c)(b-c)}\\ &=\frac{a^4b-a^4c-ab^4+b^4c+ac^4-bc^4}{(a-b)(a-c)(b-c)}\\ &=\frac{ab(a^3-b^3)+ac(c^3-a^3)+bc(b^3-c^3)}{(a-b)(a-c)(b-c)}\\ &=\frac{(a-b)(a-c)(b-c)(a^2+ab+ac+b^2+bc+c^2)}{(a-b)(a-c)(b-c)}\\ &=a^2+ab+ac+b^2+bc+c^2\\ &=\frac{2a^2+2ab+2ac+2b^2+2bc+2c^2}{2}\\ &=\frac{a^2+2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2}{2}\\ &=\frac{(a+b)^2+(a+c)^2+(b+c)^2}{2} \end{align}$$

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$$\begin{align} \small bc(b^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)&\small =bc(b^3-a^3+a^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)\\ &\small =bc(b^3-a^3)+bc(a^3-c^3)+ac(c^3-a^3)+ab(a^3-b^3)\\ &\small =-bc(a^3-b^3)+bc(a^3-c^3)-ac(a^3-c^3)+ab(a^3-b^3)\\ &\small =b(a-c)(a^3-b^3)+c(b-a)(a^3-c^3)\\ &\small =b(a-c)(a^3-b^3)-c(a-b)(a^3-c^3)\\ &\small =b(a-b)(a-c)(a^2+ab+b^2)-c(a-b)(a-c)(a^2+ac+c^2)\\ &\small =(a-b)(a-c)[b(a^2+ab+b^2)-c(a^2+ac+c^2)]\\ &\small =(a-b)(a-c)(a^2b+ab^2+b^3-a^2c-ac^2-c^3)\\ &\small =(a-b)(a-c)[a^2(b-c)+(b^3-c^3)+a(b^2-c^2)]\\ &\small =(a-b)(a-c)[a^2(b-c)+(b-c)(b^2+bc+c^2)+a(b-c)(b+c)]\\ &\small =(a-b)(a-c)(b-c)(a^2+ab+ac+b^2+bc+c^2) \end{align}$$

Answer 1

Hint: To make the factor $(a-b)(b-c)(c-a)$ appear, write $$b^3-c^3=b^3-a^3+ a^3-c^3.$$

Answer 2

Hint: Multiply $a^2+ab+ac+b^2+bc+c^2$ by $\dfrac{(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}$