If `a+b+c=5,a^(2)+b^(2)+c^(2)=12` and `a^(3)+b^(3)+c^(3)=25`. Then the value of `a^(4)+b^(4)+c^(4)` is

Options

A. `251/6`
B. `253/6`
C. `255/6`
D. `257/6`

Answer

To find the value of \( a^4 + b^4 + c^4 \) given the equations:

1. \( a + b + c = 5 \)
2. \( a^2 + b^2 + c^2 = 12 \)
3. \( a^3 + b^3 + c^3 = 25 \)

we can use the relationships between the sums of powers of \( a, b, c \).

### Step 1: Find \( ab + bc + ca \)

We can use the identity:

\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) \]

Substituting the known values:

\[ 12 = 5^2 - 2(ab + bc + ca) \]

Calculating \( 5^2 \):

\[ 12 = 25 - 2(ab + bc + ca) \]

Rearranging gives:

\[ 2(ab + bc + ca) = 25 - 12 = 13 \]

Thus,

\[ ab + bc + ca = \frac{13}{2} \]

### Step 2: Find \( abc \)

Using the identity for the sum of cubes:

\[ a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc \]

Substituting the known values:

\[ 25 = 5 \left( 12 - \frac{13}{2} \right) + 3abc \]

Calculating \( 12 - \frac{13}{2} \):

\[ 12 = \frac{24}{2} \Rightarrow 12 - \frac{13}{2} = \frac{24}{2} - \frac{13}{2} = \frac{11}{2} \]

Substituting back:

\[ 25 = 5 \cdot \frac{11}{2} + 3abc \]

Calculating \( 5 \cdot \frac{11}{2} \):

\[ 25 = \frac{55}{2} + 3abc \]

Rearranging gives:

\[ 3abc = 25 - \frac{55}{2} = \frac{50}{2} - \frac{55}{2} = -\frac{5}{2} \]

Thus,

\[ abc = -\frac{5}{6} \]

### Step 3: Find \( a^4 + b^4 + c^4 \)

Using the identity:

\[ a^4 + b^4 + c^4 = (a^2 + b^2 + c^2)^2 - 2(a^2b^2 + b^2c^2 + c^2a^2) \]

We already have \( a^2 + b^2 + c^2 = 12 \), so:

\[ (a^2 + b^2 + c^2)^2 = 12^2 = 144 \]

Now we need to find \( a^2b^2 + b^2c^2 + c^2a^2 \). We can use:

\[ a^2b^2 + b^2c^2 + c^2a^2 = (ab + bc + ca)^2 - 2abc(a + b + c) \]

Substituting the known values:

\[ = \left(\frac{13}{2}\right)^2 - 2\left(-\frac{5}{6}\right)(5) \]

Calculating \( \left(\frac{13}{2}\right)^2 \):

\[ = \frac{169}{4} \]

Calculating \( -2 \cdot -\frac{5}{6} \cdot 5 \):

\[ = \frac{10}{6} \cdot 5 = \frac{50}{6} = \frac{25}{3} \]

Now we need a common denominator to combine \( \frac{169}{4} \) and \( \frac{25}{3} \):

The common denominator of 4 and 3 is 12:

\[ \frac{169}{4} = \frac{507}{12} \]
\[ \frac{25}{3} = \frac{100}{12} \]

Thus,

\[ a^2b^2 + b^2c^2 + c^2a^2 = \frac{507}{12} + \frac{100}{12} = \frac{607}{12} \]

### Step 4: Substitute back to find \( a^4 + b^4 + c^4 \)

Now substituting back into the equation for \( a^4 + b^4 + c^4 \):

\[ a^4 + b^4 + c^4 = 144 - 2 \cdot \frac{607}{12} \]

Calculating \( 2 \cdot \frac{607}{12} = \frac{1214}{12} \):

Now we need to subtract:

\[ a^4 + b^4 + c^4 = 144 - \frac{1214}{12} \]

Converting 144 to have a denominator of 12:

\[ 144 = \frac{1728}{12} \]

Thus,

\[ a^4 + b^4 + c^4 = \frac{1728 - 1214}{12} = \frac{514}{12} = \frac{257}{6} \]

### Final Answer

The value of \( a^4 + b^4 + c^4 \) is \( \frac{257}{6} \).

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