If ` a = b cos "" ( 2 pi)/ (3) = c cos "" (4 pi)/( 3)`, then write the value of ab + bc + ca

Answer

To solve the problem, we start with the given equations:

1. \( a = b \cos\left(\frac{2\pi}{3}\right) \)
2. \( a = c \cos\left(\frac{4\pi}{3}\right) \)

We can set all of these equal to a constant \( k \):

\[ a = k, \quad b \cos\left(\frac{2\pi}{3}\right) = k, \quad c \cos\left(\frac{4\pi}{3}\right) = k \]

From these equations, we can express \( b \) and \( c \) in terms of \( k \):

\[ b = \frac{k}{\cos\left(\frac{2\pi}{3}\right)}, \quad c = \frac{k}{\cos\left(\frac{4\pi}{3}\right)} \]

Next, we need to find the values of \( \cos\left(\frac{2\pi}{3}\right) \) and \( \cos\left(\frac{4\pi}{3}\right) \):

- \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \)
- \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \)

Now substituting these values into the equations for \( b \) and \( c \):

\[ b = \frac{k}{-\frac{1}{2}} = -2k, \quad c = \frac{k}{-\frac{1}{2}} = -2k \]

Now we have:

\[ a = k, \quad b = -2k, \quad c = -2k \]

Next, we need to calculate \( ab + bc + ca \):

\[ ab = k(-2k) = -2k^2 \]
\[ bc = (-2k)(-2k) = 4k^2 \]
\[ ca = (-2k)(k) = -2k^2 \]

Now, adding these together:

\[ ab + bc + ca = -2k^2 + 4k^2 - 2k^2 = 0 \]

Thus, the value of \( ab + bc + ca \) is:

\[ \boxed{0} \]