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# hard algebra

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Simplify $$\dfrac { { a }^{ 3 }(b+c) }{ (a-b)(a-c) } +\dfrac { { b }^{ 3 }(a+c) }{ (b-a)(b-c) } +\dfrac { { c }^{ 3 }(b+a) }{ (c-b)(c-a) }$$

Jun 15, 2020

#1
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deleted

Jun 15, 2020
edited by thelizzybeth  Jun 15, 2020
#2
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Notice that

$$\dfrac { { a }^{ 3 }(b+c) }{ (a-b)(a-c) } +\dfrac { { b }^{ 3 }(a+c) }{ (b-a)(b-c) } +\dfrac { { c }^{ 3 }(b+a) }{ (c-b)(c-a) } = \dfrac{a^3(b + c)(c- b) + b^3(a + c)(a - c) + c^3(b + a)(b - a)}{(a - b)(b - c)(c - a)}$$

Expanding the numerator gives $$a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)$$.

Now, notice that the numerator is a homogeneous polynomial of degree 5. We let $$f(a, b, c) = a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)$$, which is the numerator.

We consider $$f(a, a, c) = a^3(c^2 - a^2) + a^3 (a^2 - c^2) + c^3 (a^2- a^2)$$.

$$\quad f(a, a, c) \\ = a^3(c^2 - a^2) + a^3 (a^2 - c^2) + c^3 (a^2- a^2)\\ = a^3 (c^2 - a^2 + a^2 - c^2)\\ = 0$$

By factor theorem, (a - b) is a factor of f(a, b, c).

By properties of homogeneous polynomials, (b - c) and (c - a) are also factors of f(a, b, c).

Now, consider f(a, b, c) again, and we try to factor out these common factors we just found.

$$\quad f(a, b, c) \\= a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)\\ = a^3 c^2 - a^3 b^2 + a^2 b^3 - b^3 c^2 + b^2 c^3 - a^2 c^3\\ = a^2 c^2 (a - c)+ b^3(a^2 - c^2) + b^2 (c^3 - a^3)\\ = a^2 c^2 (a - c) + b^3 (a + c)(a - c) - b^2 (a - c)(a^2 + ac + c^2)\\ = (a - c)(a^2 c^2 +b^3(a + c) - b^2(a^2 + ac +c^2))\\ = (a - c)(a^2 c^2 + ab^3 + b^3c - a^2b^2 - ab^2c - b^2 c^2)\\ = (a - c)(ab^2(b - c) + b^2c(b - c) + a^2 (c^2 - b^2))\\ = (a - c)(b - c)(ab^2 + b^2c - a^2(b + c))\\ = (a - c)(b - c)(ab(b - a)+ c(b^2 - a^2))\\ = (a - c)(b - c)(a - b)(-ab-c(a + b))\\ = (a - b)(b - c)(c - a)(ab + bc + ca)$$

Therefore, the original expression is $$\dfrac{f(a, b, c)}{(a - b)(b - c)(c - a)} = \dfrac{(a - b)(b - c)(c - a)(ab + bc + ca)}{(a - b)(b - c)(c - a)} = \boxed{ab + bc + ca}$$

Jun 15, 2020
edited by MaxWong  Jun 15, 2020
#3
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The amount of work to get the answer is dazzling.

Jun 15, 2020
#4
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Simplify

$$\dfrac { { a }^{ 3 }(b+c) }{ (a-b)(a-c) } +\dfrac { { b }^{ 3 }(a+c) }{ (b-a)(b-c) } +\dfrac { { c }^{ 3 }(b+a) }{ (c-b)(c-a) }$$

$$\begin{array}{|rcll|} \hline && \mathbf{\dfrac {a^3(b+c) }{ (a-b)(a-c) } +\dfrac {b^3(a+c)}{ (b-a)(b-c) } +\dfrac {c^3(b+a)}{ (c-b)(c-a) } } \\\\ &=& \dfrac {a^3(b+c) }{ (a-b)(a-c) } -\dfrac {b^3(a+c)}{ (a-b)(b-c) } +\dfrac {c^3(a+b)}{ (b-c)(a-c) } \\\\ &=& \dfrac {a^3(b+c)(b-c)- b^3(a+c)(a-c)+c^3(a+b)(a-b)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3(b+c)(b-c)+c^3(a+b)(a-b)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3(b^2-c^2)+c^3(a^2-b^2)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3b^2-a^3c^2+c^3a^2-c^3b^2- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3b^2-c^3b^2-a^3c^2+c^3a^2- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {b^2(a^3-c^3)-a^2c^2(a-c)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {b^2(a-c)(a^2+ac+c^2)-a^2c^2(a-c)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {(a-c)\left(b^2(a^2+ac+c^2)-a^2c^2- b^3(a+c)\right)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac { b^2(a^2+ac+c^2)-a^2c^2- b^3(a+c) }{ (a-b)(b-c) } \\\\ &=& \dfrac { b^2a^2+b^2ac+b^2c^2 -a^2c^2- b^3a-b^3c }{ (a-b)(b-c) } \\\\ &=& \dfrac { b^2a^2- b^3a +b^2ac-b^3c +b^2c^2 -a^2c^2 }{ (a-b)(b-c) } \\\\ &=& \dfrac { ab^2(a- b) +b^2c(a-b) -c^2( a^2-b^2) }{ (a-b)(b-c) } \\\\ &=& \dfrac { (ab^2+b^2c)(a- b)-c^2(a+b)(a-b) }{ (a-b)(b-c) } \\\\ &=& \dfrac {(a-b)\left( (ab^2+b^2c)-c^2(a+b)\right) }{ (a-b)(b-c) } \\\\ &=& \dfrac {ab^2+b^2c -c^2(a+b)}{(b-c) } \\\\ &=& \dfrac {ab^2+b^2c -c^2a-c^2b}{(b-c) } \\\\ &=& \dfrac {ab^2 -c^2a +b^2c-c^2b}{(b-c) } \\\\ &=& \dfrac {a(b^2 -c^2) +bc(b-c)}{(b-c) } \\\\ &=& \dfrac {a(b+c)(b-c) +bc(b-c)}{(b-c) } \\\\ &=& \dfrac {(b-c)\Big(a(b+c) +bc\Big)}{(b-c) } \\\\ &=& a(b+c) +bc \\ &=& \mathbf{ab+ac+bc} \\ \hline \end{array}$$

Jun 16, 2020