# Show that |(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|=-2(a^3+b^3+c^3-3abc?

Aug 6, 2018

#### Explanation:

We know that ,
${x}^{3} + {y}^{3} + {z}^{3}$=(x+y+z)(x^2+${y}^{2}$+z^2-xy-yz-zx)+3xyz
Let ,
$D = | \left(a + b , b + c , c + a\right) , \left(b + c , c + a , a + b\right) , \left(c + a , a + b , b + c\right) |$
Taking ,${R}_{1} + {R}_{2} + {R}_{3}$

$D = | \left(2 a + 2 b + 2 c , 2 a + 2 b + 2 c , 2 a + 2 b + 2 c\right) , \left(b + c , c + a , a + b\right) , \left(c + a , a + b , b + c\right) |$

Taking ${R}_{1} \left(\frac{1}{2 a + 2 b + 2 c}\right)$

$D = \left(2 a + 2 b + 2 c\right) | \left(1 , 1 , 1\right) , \left(b + c , c + a , a + b\right) , \left(c + a , a + b , b + c\right) |$

Use , ${C}_{2} - {C}_{1} \mathmr{and} {C}_{3} - {C}_{2}$

$D = 2 \left(a + b + c\right) | \left(1 , 1 - 1 , 1 - 1\right) , \left(b + c , c + a - b - c , a + b - c - a\right) , \left(c + a , a + b - c - a , b + c - a - b\right) |$

$\therefore D = 2 \left(a + b + c\right) | \left(1 , 0 , 0\right) , \left(b + c , a - b , b - c\right) , \left(c + a , b - c , c - a\right) |$

Expanding we get

$D = 2 \left(a + b + c\right) \left\{1 \left(a - b\right) \left(c - a\right) - \left(b - c\right) \left(b - c\right) - 0 = 0\right\}$

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

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

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

Using above formula for ${x}^{3} + {y}^{3} + {z}^{3}$

$\therefore D = - 2 \left({a}^{3} + {b}^{3} + {c}^{3} - 3 a b c\right)$