# How do you simplify (a+4b-3c)^3?

Jul 27, 2015

a^3 + 12 a^2 b + 48 a b^2 + 64 b^3 - 9 a^2 c - 72 a b c - 144 b^2 c + 27 a c^2 + 108 b c^2 - 27 c^3

#### Explanation:

Take it step-by-step, using the fact that when you multiply one polynomial times another, each of the terms in the first one needs to get multiplied one time by each of the terms in the second one (this follows from the distributive property).

So, first we compute ${\left(a + 4 b - 3 c\right)}^{2}$:

${\left(a + 4 b - 3 c\right)}^{2} = \left(a + 4 b - 3 c\right) \cdot \left(a + 4 b - 3 c\right)$

$= {a}^{2} + 4 a b - 3 a c + 4 a b + 16 {b}^{2} - 12 b c - 3 a c - 12 b c + 9 {c}^{2}$

Combine like terms to get:

${\left(a + 4 b - 3 c\right)}^{2} = {a}^{2} + 16 {b}^{2} + 9 {c}^{2} + 8 a b - 6 a c - 24 b c$

Now

${\left(a + 4 b - 3 c\right)}^{3} = \left(a + 4 b - 3 c\right) \cdot \left({a}^{2} + 16 {b}^{2} + 9 {c}^{2} + 8 a b - 6 a c - 24 b c\right)$

$= {a}^{3} + 16 a {b}^{2} + 9 a {c}^{2} + 8 {a}^{2} b - 6 {a}^{2} c - 24 a b c + 4 {a}^{2} b + 64 {b}^{3} + 36 b {c}^{2} + 32 a {b}^{2} - 24 a b c - 96 {b}^{2} c - 3 {a}^{2} c - 48 {b}^{2} c - 27 {c}^{3} - 24 a b c + 18 a {c}^{2} + 72 b {c}^{2}$

=a^3 + 12 a^2 b + 48 a b^2 + 64 b^3 - 9 a^2 c - 72 a b c - 144 b^2 c + 27 a c^2 + 108 b c^2 - 27 c^3

Jul 27, 2015

Look at general case ${\left(A + B + C\right)}^{3}$ which is easier to handle, then substitute $A = a$, $B = 4 b$ and $C = - 3 c$ to find:

${\left(a + 4 b - 3 c\right)}^{3}$

$= {a}^{3} + 64 {b}^{3} - 27 {c}^{3} + 12 {a}^{2} b - 144 {b}^{2} c + 27 {c}^{2} a - 9 {a}^{2} c + 48 {b}^{2} a + 108 {c}^{2} b - 72 a b c$

#### Explanation:

Consider the general case of the cube of a trinomial:

${\left(A + B + C\right)}^{3} = \left(A + B + C\right) \left(A + B + C\right) \left(A + B + C\right)$

This is easier to solve than our original problem because it is symmetrical in $A$, $B$ and $C$.

The only way we can get a multiple of ${A}^{3}$ is by picking the $A$ from each of the three trinomials. This can only be done in one way, so the coefficient of ${A}^{3}$ is $1$. Similarly, the coefficient of ${B}^{3}$ and ${C}^{3}$ must be $1$.

We can get a multiple of ${A}^{2} B$ by picking one of the $3$ trinomials to take our $B$ from then take $A$ from the other two. Since we can arrive at ${A}^{2} B$ in $3$ ways, that is the coefficient of ${A}^{2} B$, ${B}^{2} C$, ${C}^{2} A$, ${A}^{2} C$, ${B}^{2} A$ and ${C}^{2} B$.

We can get a multiple of $A B C$ by picking one of the $3$ trinomials to take $A$ from, then one of the $2$ remaining trinomials to take $B$ from, leaving us with one trinomial to take $C$ from. So there are a total of $3 \times 2 = 6$ ways to do this.

So:
${\left(A + B + C\right)}^{3}$
$= {A}^{3} + {B}^{3} + {C}^{3} + 3 {A}^{2} B + 3 {B}^{2} C + 3 {C}^{2} A + 3 {A}^{2} C + 3 {B}^{2} A + 3 {C}^{2} B + 6 A B C$

Now let $A = a$, $B = 4 b$ and $C = - 3 c$ to find:

${\left(a + 4 b - 3 c\right)}^{3}$

$= {\left(A + B + C\right)}^{3}$

$= {A}^{3} + {B}^{3} + {C}^{3} + 3 {A}^{2} B + 3 {B}^{2} C + 3 {C}^{2} A + 3 {A}^{2} C + 3 {B}^{2} A + 3 {C}^{2} B + 6 A B C$

$= {a}^{3} + 64 {b}^{3} - 27 {c}^{3} + 12 {a}^{2} b - 144 {b}^{2} c + 27 {c}^{2} a - 9 {a}^{2} c + 48 {b}^{2} a + 108 {c}^{2} b - 72 a b c$