Simiplifying a Complicated Expression

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High School Math Grades 9-12, Ages 13-18

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Simiplifying a Complicated Expression

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#1 h Mar 15, 2010, 3:59 AM • 2 Y

Y by Adventure10 and 1 other user

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

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#2 h Mar 15, 2010, 4:06 AM • 2 Y

Y by Adventure10, Mango247

There are too many steps but I am getting a+b+c.

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#3 h Mar 15, 2010, 4:59 PM • 2 Y

Y by Adventure10, Mango247

You could use the fact that the fractions in the expression are symmetric?

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#4 h Mar 16, 2010, 5:54 PM • 2 Y

Y by Adventure10 and 1 other user

After expanding the expression, just divide the numerator by the denominator and we will get a + b + c.

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#5 h Mar 16, 2010, 9:50 PM • 1 Y

Y by Adventure10

If we combine it into one fraction we get $\frac{a^3b-ab^3+b^3c-bc^3+c^3a-ca^3}{(a-b)(a-c)(b-c)}$ . But in the numerator if $a=b$ then the numerator would would $0$ , thus $a-b$ must be a factor of the numerator, similarly $a-c$ and $b-c$ must be also, dividing them out we get $a+b+c$ .

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#6 h Apr 22, 2025, 12:47 AM

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Let $\frac{a^3}{(a - b)(a - c)} + \frac{b^3}{(b - c)(b - a)} + \frac{c^3}{(c - a)(c - b)} = t$
$t = -\frac{a^3(b - c) + b^3(c - a) + c^3(a - b)}{(a - b)(b - c)(c - a)}$
Now, $N_r = a^3(b - c) + b^3(c - a) + c^3(a - b)$
$= a^3 b - ab^3 + b^3c - ca^3 + c^3a - bc^3$
$= ab(a^2 - b^2) - c(a^3 - b^3) + c^3(a - b)$
$= (a - b)[ab(a + b) - c(a^2 + ab + b^2) + c^3]$
$= (a - b)[a²b + ab² - ca² - abc - b²c + c³]$
$= (a - b)[a²(b - c) + ab(b - c) - c(b² - c²)]$
$= (a - b)(b - c)[a² + ab - c(b + c)]$
$= (a - b)(b - c)[a² + ab - bc - c²)]$
$= (a - b)(b - c)[-b(c - a) - (c² - a²)]$
$= -(a - b)(b - c)(c - a)(a + b + c)$
Therefore, $t = \frac{(a - b)(b - c)(c - a)(a + b + c)}{(a - b)(b - c)(c - a)} = \boxed{a + b + c}$

This post has been edited 1 time. Last edited by P162008, Apr 22, 2025, 12:49 AM
Reason: Typo

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#7 h Apr 23, 2025, 5:35 AM

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This is the coefficient of the $x^2$ term in $x^3 - (x-a)(x-b)(x-c)$ by Lagrange interpolation.

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