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Question 1209716: Let a, b, c be the roots of p(x) = x^3 + 7x^2 + 10x - 13 - 5x^3 + 25x^2 + 44x - 15. Find the value of
1/(ab + ac) + 1/(ab + bc) + 1/(ac + bc)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
1. **Simplify the polynomial:**
p(x) = x³ + 7x² + 10x - 13 - 5x³ + 25x² + 44x - 15
p(x) = -4x³ + 32x² + 54x - 28
2. **Relate roots to coefficients:**
For a cubic polynomial ax³ + bx² + cx + d = 0 with roots a, b, and c, we have the following relationships:
* a + b + c = -b/a
* ab + ac + bc = c/a
* abc = -d/a
In our case, -4x³ + 32x² + 54x - 28 = 0. Dividing by -4 gives
x³ - 8x² - (27/2)x + 7 = 0
So,
* a + b + c = 8
* ab + ac + bc = -27/2
* abc = -7
3. **Simplify the expression:**
We want to find the value of:
1/(ab + ac) + 1/(ab + bc) + 1/(ac + bc)
We can rewrite this as:
1/[a(b + c)] + 1/[b(a + c)] + 1/[c(a + b)]
Find a common denominator:
[b(a + c)c(a + b) + a(b + c)c(a + b) + a(b + c)b(a + c)] / [a(b + c)b(a + c)c(a + b)]
Expand the numerator:
[abc + bc² + abc + b²c + a²c + ac² + a²b + ab² + abc + bc² + abc + b²c] / [abc(a + b)(b + c)(c + a)]
Simplify the numerator:
[2abc + bc² + b²c + a²c + ac² + a²b + ab²] / [abc(a + b)(b + c)(c + a)]
Notice that the numerator can be rearranged as:
[ab(a + b) + bc(b + c) + ac(a + c) + 2abc]
4. **Substitute the relationships:**
We have:
1/(ab + ac) + 1/(ab + bc) + 1/(ac + bc) = [ab(a+b) + bc(b+c) + ac(a+c) + 2abc]/[abc(a+b)(b+c)(c+a)]
We also know that (a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) - abc, so (a+b)(b+c)(c+a) = 8*(-27/2) - (-7) = -108 + 7 = -101
ab(a+b) + bc(b+c) + ac(a+c) = (a+b+c)(ab+bc+ac) - 3abc = 8*(-27/2) - 3(-7) = -108 + 21 = -87
Therefore, the expression becomes:
(-87 + 2(-7)) / (-7 * -101) = (-87 - 14) / 707 = -101/707 = -1/7
Therefore, 1/(ab + ac) + 1/(ab + bc) + 1/(ac + bc) = -1/7.
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