+658 Group every two terms and factor (a,b,c) respectively:
\(a(b^3-a^2b)+b(c^3-b^2c)+c(a^3-c^2a)\)
Now factor (b,c,a) respectively:
\(ab(b^2-a^2)+bc(c^2-b^2)+ca(a^2-c^2)\)
Now obviously follow up to factor the subtraction of two perfect squares.
\(ab(b-a)(b+a)+bc(c-b)(c+b)+ca(a-c)(a+c)\)
Not sure if this can be factored further
+26393 Factor \(ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a.\)
\(\begin{array}{|rcll|} \hline && \mathbf{ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a} \\ &=& -a^3 b+ ab^3 + ca^3 - b^3 c - c^3 a + bc^3 \\ &=& -ab(a^2-b^2) + c(a^3 - b^3) - c^3( a -b) \\ &=& -ab(a-b)(a+b) + c(a-b)(a^2+ab+b^2) - c^3( a -b) \\ &=& (a-b)\Big( -ab(a+b) + c(a^2+ab+b^2) - c^3 \Big) \\ &=& (a-b)( -a^2b-ab^2 + ca^2+cab+cb^2 - c^3) \\ &=& (a-b)( ca^2 -a^2b +cab -ab^2 - c^3 +cb^2 ) \\ &=& (a-b)\Big( a^2(c-b) + ab(c-b)-c(c^2-b^2) \Big) \\ &=& (a-b)\Big( a^2(c-b) + ab(c-b)-c(c-b)(c+b) \Big) \\ &=& (a-b)(c-b)\Big( a^2 + ab-c(c+b) \Big) \\ &=& (a-b)(c-b)(a^2 + ab-c^2-cb) \\ &=& (a-b)(c-b)(a^2-c^2 + ab-cb) \\ &=& (a-b)(c-b)\Big((a-c)(a+c) + b(a-c) \Big) \\ &=& (a-b)(c-b)(a-c)(a+c+b) \\ &=& \mathbf{(a-b)(c-b)(a-c)(a+b+c)} \\ \hline \end{array}\)
