# The question is below?

## Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x = c y + b z , y = a z + c x \mathmr{and} z = b x + a y$ then find the value of ${a}^{2} + {b}^{2} + {c}^{2}$.

Jun 4, 2018

$1 - 2 a b c$.

#### Explanation:

We first eliminate $z$.

$x = c y + b z , \text{ where, } z = b x + a y$

$\therefore x = c y + b \left(b x + a y\right) = c y + {b}^{2} x + a b y$.

$\therefore x \left(1 - {b}^{2}\right) = y \left(c + a b\right)$

$\therefore \frac{x}{y} = \frac{c + a b}{1 - {b}^{2}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(1\right)$.

Similarly, $y = a z + c x = a \left(b x + a y\right) + c x = a b x + {a}^{2} y + c x$,

$\therefore y \left(1 - {a}^{2}\right) = x \left(c + a b\right)$.

$\therefore \frac{x}{y} = \frac{1 - {a}^{2}}{c + a b} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(2\right)$.

$\left(1\right) \mathmr{and} \left(2\right) \Rightarrow \frac{c + a b}{1 - {b}^{2}} = \frac{x}{y} = \frac{1 - {a}^{2}}{c + a b}$,

$i . e . , {\left(c + a b\right)}^{2} = \left(1 - {a}^{2}\right) \left(1 - {b}^{2}\right)$.

$\therefore {c}^{2} + 2 a b c + {a}^{2} {b}^{2} = 1 - {a}^{2} - {b}^{2} + {a}^{2} {b}^{2}$.

$\therefore {a}^{2} + {b}^{2} + {c}^{2} = 1 - 2 a b c$, is the desired value!