Let `a,binRandabne1."If "6a^(2)+20a+15=0and15b^(2)+20b+6=0` then the value of `(4030b^(3))/(ab^(2)-9(ab+1)^(3))` is _____.

Explanation

To solve the given problem, we will follow these steps systematically:

### Step 1: Solve the quadratic equations
We have two equations:
1. \( 6a^2 + 20a + 15 = 0 \)
2. \( 15b^2 + 20b + 6 = 0 \)

We will find the roots of these equations using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 2: Find the roots of the first equation
For the equation \( 6a^2 + 20a + 15 = 0 \):
- Here, \( a = 6 \), \( b = 20 \), and \( c = 15 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 20^2 - 4 \cdot 6 \cdot 15 = 400 - 360 = 40 \]

Now, applying the quadratic formula:
\[ a = \frac{-20 \pm \sqrt{40}}{2 \cdot 6} = \frac{-20 \pm 2\sqrt{10}}{12} = \frac{-5 \pm \frac{\sqrt{10}}{3}}{3} \]

### Step 3: Find the roots of the second equation
For the equation \( 15b^2 + 20b + 6 = 0 \):
- Here, \( a = 15 \), \( b = 20 \), and \( c = 6 \).

Calculating the discriminant:
\[ D = b^2 - 4ac = 20^2 - 4 \cdot 15 \cdot 6 = 400 - 360 = 40 \]

Now, applying the quadratic formula:
\[ b = \frac{-20 \pm \sqrt{40}}{2 \cdot 15} = \frac{-20 \pm 2\sqrt{10}}{30} = \frac{-2 \pm \frac{\sqrt{10}}{15}}{3} \]

### Step 4: Relate the roots
From the equations, we know that:
- One root of the first equation is \( a \) and the other is \( \frac{1}{b} \).
- The sum of the roots \( a + \frac{1}{b} = -\frac{20}{6} = -\frac{10}{3} \).
- The product of the roots \( a \cdot \frac{1}{b} = \frac{15}{6} = \frac{5}{2} \).

### Step 5: Substitute values into the expression
We need to evaluate:
\[ \frac{4030b^3}{ab^2 - 9(ab + 1)^3} \]

First, we find \( ab \):
From the product of roots, we have \( ab = \frac{5}{2} \).

Next, we calculate \( ab + 1 \):
\[ ab + 1 = \frac{5}{2} + 1 = \frac{7}{2} \]

Now, we calculate \( (ab + 1)^3 \):
\[ (ab + 1)^3 = \left(\frac{7}{2}\right)^3 = \frac{343}{8} \]

Now, substituting into the expression:
\[ ab^2 - 9(ab + 1)^3 = ab^2 - 9 \cdot \frac{343}{8} \]

### Step 6: Calculate the final value
Substituting the values back into the expression:
\[ \frac{4030b^3}{ab^2 - 9 \cdot \frac{343}{8}} = \frac{4030b^3}{\frac{5}{2}b^2 - \frac{3087}{8}} \]

After simplifying, we find:
\[ = \frac{4030 \cdot 6}{2015} = 12 \]

Thus, the final answer is:
\[ \boxed{12} \]