# a + 2c + 3b = 0 then what is  a ^3 + 8 b ^3 + 27c^3?

Feb 18, 2015

We can rewrite one of the letters as an expression in the others.

$a + 2 c + 3 b = 0 \to a = - 2 c - 3 b$

Then we replace every $a$ in the second expression:

${a}^{3} + 8 {b}^{3} + 27 {c}^{3} =$
${\left(- 2 c - 3 b\right)}^{3} + 8 {b}^{3} + 27 {c}^{3} =$
$\left(- 8 {c}^{3} - 36 b {c}^{2} - 54 {b}^{2} c - 19 {b}^{3}\right) + 8 {b}^{3} + 27 {c}^{3} =$

$19 {c}^{3} - 36 b {c}^{2} - 54 {b}^{2} c - 10 {b}^{3}$

Extra :
If the first expression had been $a + 2 b + 3 c = 0$
(and I suspect that it was!!)

Then the ${b}^{3}$'s and ${c}^{3}$'s would have cancelled out nicer.

$a = - 2 b - 3 c$
${\left(- 2 b - 3 c\right)}^{3} + 8 {b}^{3} + 27 {c}^{3} =$
$\left(- 8 {b}^{3} - 36 {b}^{2} c - 54 b {c}^{2} - 27 {c}^{3}\right) + 8 {b}^{3} + 27 {c}^{3} =$

$- 36 {b}^{2} c - 54 b {c}^{2} \mathmr{and} 18 b c \left(- 2 b - 3 c\right)$
and, since $a = - 2 b - 3 c$ we can rewrite that:

${a}^{3} + 8 {b}^{3} + 27 {c}^{3} = 18 a b c$