If the equation x^2 + ax + b = 0 where a, b belongs to R has a non real root whose cube is 343 then (7a+b) has the value

🧠 Problem Statement
If the equation 
$$x^2 + ax + b = 0$$
where a, b belongs to R has a non real root whose cube is 343 then (7a+b) has the value

🔍 Step 1: Find the Cube Roots of 343
The equation $x^3 = 343$ can be rewritten as:
$$x^3 - 7^3 = 0$$
Factoring this, we get:
$$(x - 7)(x^2 + 7x + 49) = 0$$
Thus, the roots are:
$x = 7$ (real root)
$x=7\mathrm{\omega \; and}$$x = 7\omega^2$ (non-real roots), where $\omega$ is a cube root of unity.

The cube roots of unity are:
$$\omega = \frac{-1 + i\sqrt{3}}{2}, \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2}$$
Therefore, the non-real cube roots of 343 are:
$$7\omega = 7 \cdot \frac{-1 + i\sqrt{3}}{2}, \quad 7\omega^2 = 7 \cdot \frac{-1 - i\sqrt{3}}{2}$$
🧩 Step 2: Determine the Quadratic Equation
Since the original quadratic equation has real coefficients and a non-real root, its roots must be complex conjugates. Let's denote the roots as:
$$x = 7\omega \quad \text{and} \quad x = 7\omega^2$$
Using the fact that for a quadratic equation $x^2 + ax + b = 0$, the sum and product of the roots are:
$$\text{Sum of roots} = -a = 7\omega + 7\omega^2 = 7(\omega + \omega^2)$$


$$\text{Product of roots} = b = 7\omega \cdot 7\omega^2 = 49\omega^3$$
We know that:
$$\omega + \omega^2 = -1, \quad \omega^3 = 1$$
Therefore:
$$-a = 7(-1) \Rightarrow a = 7$$
$$b = 49 \cdot 1 = 49$$

✅ Final Answer
The values of $a$ and b are
$$a = 7, \quad b = 49$$
Thus, the value of $7a + b$ is:
$$7a + b = 7 \cdot 7 + 49 = 49 + 49 = \boxed{98}$$