If the equation `x^(5)-10a^(3)x^(2)+b^(4)x+c^(5)=0` has three equal roots, then

Options

A. `2b^(2)-10a^(3)b^(2)+c^(5)=0`
B. `6a^(2)+c^(5)=0`
C. `2x^(5)-10a^(3)b^(2)+b^(4)c^(5)=0`
D. `b^(4)=15a^(4)`

Answer

To solve the problem, we need to analyze the given polynomial equation:

\[ x^5 - 10a^3x^2 + b^4x + c^5 = 0 \]

We are told that this equation has three equal roots. Let's denote the equal root as \( \alpha \). Since the equation is of degree 5, it can have up to 5 roots, and having three equal roots means that the polynomial can be expressed in the form:

\[ (x - \alpha)^3(x - r_1)(x - r_2) = 0 \]

where \( r_1 \) and \( r_2 \) are the other two roots.

### Step 1: Differentiate the polynomial

To find the conditions for the roots, we need to differentiate the polynomial. The first derivative \( f'(x) \) will help us find the critical points:

\[ f'(x) = 5x^4 - 20a^3x + b^4 \]

### Step 2: Set the first derivative to zero

Since \( \alpha \) is a root of multiplicity 3, it must also be a critical point of the polynomial. Therefore, we set the first derivative equal to zero at \( x = \alpha \):

\[ f'(\alpha) = 5\alpha^4 - 20a^3\alpha + b^4 = 0 \]

### Step 3: Differentiate again

Next, we differentiate the first derivative to find the second derivative:

\[ f''(x) = 20x^3 - 20a^3 \]

### Step 4: Set the second derivative to zero

For \( \alpha \) to be a point of inflection (which is necessary for a root of multiplicity greater than 2), we set the second derivative equal to zero:

\[ f''(\alpha) = 20\alpha^3 - 20a^3 = 0 \]

From this, we can conclude:

\[ \alpha^3 = a^3 \]
\[ \alpha = a \]

### Step 5: Substitute \( \alpha \) back into the first derivative equation

Now that we have \( \alpha = a \), we substitute this back into the equation from Step 2:

\[ 5a^4 - 20a^4 + b^4 = 0 \]
\[ -15a^4 + b^4 = 0 \]
\[ b^4 = 15a^4 \]
\[ b = \sqrt[4]{15} a \]

### Step 6: Substitute \( \alpha \) back into the original polynomial

Next, we substitute \( \alpha = a \) into the original polynomial to find the relationship with \( c \):

\[ f(a) = a^5 - 10a^5 + b^4a + c^5 = 0 \]
\[ -9a^5 + b^4a + c^5 = 0 \]

Using \( b^4 = 15a^4 \):

\[ -9a^5 + 15a^4a + c^5 = 0 \]
\[ -9a^5 + 15a^5 + c^5 = 0 \]
\[ 6a^5 + c^5 = 0 \]
\[ c^5 = -6a^5 \]
\[ c = -\sqrt[5]{6} a \]

### Conclusion

Thus, we have derived the relationships between \( a \), \( b \), and \( c \):

- \( b = \sqrt[4]{15} a \)
- \( c = -\sqrt[5]{6} a \)