Đáp án:

Giải thích các bước giải:

`A=(a^3)/((a-b)(a-c))+(b^3)/((b-a)(b-c))+(c^3)/((c-a)(c-b))`

`ĐK:a \ne b \ne c`

`A=(a^3)/((a-b)(a-c))-(b^3)/((a-b)(b-c))+(c^3)/((a-c)(b-c))`

`A=(a^3(b-c))/((a-b)(a-c)(b-c))-(b^3(a-c))/((a-b)(a-c)(b-c))+(c^3(a-b))/((a-b)(a-c)(b-c))`

`A=(a^3(b-c)-b^3(a-c)+c^3(a-b))/((a-b)(a-c)(b-c))`

`A=((a-b)(a-c)(b-c)(a+b+c))/((a-b)(a-c)(b-c))`

`A=(a+b+c)`

$\begin{array}{l} \dfrac{{{a^3}}}{{\left( {a - b} \right)\left( {a - c} \right)}} + \dfrac{{{b^3}}}{{\left( {b - c} \right)\left( {b - a} \right)}} + \dfrac{{{c^3}}}{{\left( {c - b} \right)\left( {c - a} \right)}}\\  = \dfrac{{{a^3}\left( {c - b} \right) + {b^3}\left( {a - c} \right) + {c^3}\left( {b - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\  = \dfrac{{{a^3}c - {c^3}a + {b^3}a - {a^3}b + {c^3}b - {b^3}c}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}\\ TS:\\ {a^3}c - {c^3}a + {b^3}a - {a^3}b + {c^3}b - {b^3}c\\  = ac\left( {{a^2} - {c^2}} \right) + {b^3}\left( {a - c} \right) - b\left( {{a^3} - {c^3}} \right)\\  = ac\left( {a - c} \right)\left( {a + c} \right) + {b^3}\left( {a - c} \right) - b\left( {a - c} \right)\left( {{a^2} + ac + {c^2}} \right)\\  = \left( {a - c} \right)\left( {ac\left( {a + c} \right) + {b^3} - b\left( {{a^2} + ac + {c^2}} \right)} \right)\\  = \left( {a - c} \right)\left( {{a^2}c + a{c^2} + {b^3} - {a^2}b - abc - b{c^2}} \right)\\  = \left( {a - c} \right)\left[ {ac\left( {a - b} \right) + {c^2}\left( {a - b} \right) - b\left( {{a^2} - {b^2}} \right)} \right]\\  = \left( {a - c} \right)\left( {a - b} \right)\left( {ac + {c^2} - b\left( {a + b} \right)} \right)\\  = \left( {a - c} \right)\left( {a - b} \right)\left( {ac + {c^2} - ab - {b^2}} \right)\\  = \left( {a - c} \right)\left( {a - b} \right)\left[ {a\left( {c - b} \right) + \left( {c - b} \right)\left( {c + b} \right)} \right]\\  = \left( {a - c} \right)\left( {a - b} \right)\left( {c - b} \right)\left( {a + b + c} \right)\\  = \left( {c - a} \right)\left( {b - c} \right)\left( {a - b} \right)\left( {a + b + c} \right)\\  \Rightarrow A = a + b + c \end{array}$