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Question: Let a, b, c are roots of equation $x^3 + 8x + 1 = 0$, then the value of $\frac{bc}{(8b+1)(8c+1)} + \...

Let a, b, c are roots of equation x3+8x+1=0x^3 + 8x + 1 = 0, then the value of bc(8b+1)(8c+1)+ac(8a+1)(8c+1)+ab(8a+1)(8b+1)\frac{bc}{(8b+1)(8c+1)} + \frac{ac}{(8a+1)(8c+1)} + \frac{ab}{(8a+1)(8b+1)} is equal to

A

0

B

-8

C

-16

D

16

Answer

-16

Explanation

Solution

The problem asks us to find the value of a symmetric expression involving the roots of a cubic equation.

1. Identify the given equation and its roots: The given equation is x3+8x+1=0x^3 + 8x + 1 = 0.
Let a, b, c be the roots of this equation.

2. Apply Vieta's formulas:
For a cubic equation Ax3+Bx2+Cx+D=0Ax^3 + Bx^2 + Cx + D = 0, the Vieta's formulas are:

  • Sum of roots: a+b+c=−B/Aa+b+c = -B/A
  • Sum of products of roots taken two at a time: ab+bc+ca=C/Aab+bc+ca = C/A
  • Product of roots: abc=−D/Aabc = -D/A

In our equation, x3+0x2+8x+1=0x^3 + 0x^2 + 8x + 1 = 0, so A=1,B=0,C=8,D=1A=1, B=0, C=8, D=1.
Therefore:

  • a+b+c=−0/1=0a+b+c = -0/1 = 0
  • ab+bc+ca=8/1=8ab+bc+ca = 8/1 = 8
  • abc=−1/1=−1abc = -1/1 = -1

3. Simplify the terms in the denominator:
Since a, b, c are roots of x3+8x+1=0x^3 + 8x + 1 = 0, they must satisfy the equation:

  • For root a: a3+8a+1=0  ⟹  8a+1=−a3a^3 + 8a + 1 = 0 \implies 8a+1 = -a^3
  • For root b: b3+8b+1=0  ⟹  8b+1=−b3b^3 + 8b + 1 = 0 \implies 8b+1 = -b^3
  • For root c: c3+8c+1=0  ⟹  8c+1=−c3c^3 + 8c + 1 = 0 \implies 8c+1 = -c^3

4. Substitute the simplified terms into the expression:
The expression to evaluate is:
E=bc(8b+1)(8c+1)+ac(8a+1)(8c+1)+ab(8a+1)(8b+1)E = \frac{bc}{(8b+1)(8c+1)} + \frac{ac}{(8a+1)(8c+1)} + \frac{ab}{(8a+1)(8b+1)}

Substitute the simplified denominators:
E=bc(−b3)(−c3)+ac(−a3)(−c3)+ab(−a3)(−b3)E = \frac{bc}{(-b^3)(-c^3)} + \frac{ac}{(-a^3)(-c^3)} + \frac{ab}{(-a^3)(-b^3)}
E=bcb3c3+aca3c3+aba3b3E = \frac{bc}{b^3c^3} + \frac{ac}{a^3c^3} + \frac{ab}{a^3b^3}

5. Simplify each term and find a common denominator:
E=1b2c2+1a2c2+1a2b2E = \frac{1}{b^2c^2} + \frac{1}{a^2c^2} + \frac{1}{a^2b^2}

To combine these fractions, the common denominator is (abc)2(abc)^2:
E=a2(abc)2+b2(abc)2+c2(abc)2E = \frac{a^2}{(abc)^2} + \frac{b^2}{(abc)^2} + \frac{c^2}{(abc)^2}
E=a2+b2+c2(abc)2E = \frac{a^2+b^2+c^2}{(abc)^2}

6. Use the value of abcabc from Vieta's formulas:
We found abc=−1abc = -1.
So, (abc)2=(−1)2=1(abc)^2 = (-1)^2 = 1.

Substitute this into the expression for E:
E=a2+b2+c21E = \frac{a^2+b^2+c^2}{1}
E=a2+b2+c2E = a^2+b^2+c^2

7. Calculate a2+b2+c2a^2+b^2+c^2 using an algebraic identity:
We know the identity: (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)
Rearranging to find a2+b2+c2a^2+b^2+c^2:
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ca)

Substitute the values from Vieta's formulas (a+b+c=0a+b+c=0 and ab+bc+ca=8ab+bc+ca=8):
a2+b2+c2=(0)2−2(8)a^2+b^2+c^2 = (0)^2 - 2(8)
a2+b2+c2=0−16a^2+b^2+c^2 = 0 - 16
a2+b2+c2=−16a^2+b^2+c^2 = -16

Therefore, the value of the expression EE is −16-16.