If a+b =`7/3` and `a^(2) + b^(2) = 31/9` , find `27(a^(3) + b^(3))`

Options

A. A)154
B. B)156
C. C)152
D. D)164

Answer

To solve the problem, we need to find \(27(a^3 + b^3)\) given that \(a + b = \frac{7}{3}\) and \(a^2 + b^2 = \frac{31}{9}\).

### Step-by-step Solution:

**Step 1: Use the identity for \(a^2 + b^2\)**
We know that:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Substituting the values we have:
\[ \frac{31}{9} = \left(\frac{7}{3}\right)^2 - 2ab \]
Calculating \(\left(\frac{7}{3}\right)^2\):
\[ \left(\frac{7}{3}\right)^2 = \frac{49}{9} \]
So, we can write:
\[ \frac{31}{9} = \frac{49}{9} - 2ab \]
Rearranging gives us:
\[ 2ab = \frac{49}{9} - \frac{31}{9} = \frac{18}{9} = 2 \]
Thus, we find:
\[ ab = 1 \]

**Step 2: Use the identity for \(a^3 + b^3\)**
We know that:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
We can express \(a^2 - ab + b^2\) as:
\[ a^2 - ab + b^2 = (a^2 + b^2) - ab \]
Substituting the known values:
\[ a^2 - ab + b^2 = \frac{31}{9} - 1 = \frac{31}{9} - \frac{9}{9} = \frac{22}{9} \]
Now substituting back into the formula for \(a^3 + b^3\):
\[ a^3 + b^3 = (a + b)\left(a^2 - ab + b^2\right) = \left(\frac{7}{3}\right)\left(\frac{22}{9}\right) \]
Calculating this gives:
\[ a^3 + b^3 = \frac{7 \times 22}{3 \times 9} = \frac{154}{27} \]

**Step 3: Calculate \(27(a^3 + b^3)\)**
Now we multiply by 27:
\[ 27(a^3 + b^3) = 27 \times \frac{154}{27} = 154 \]

Thus, the final answer is:
\[ \boxed{154} \]