Think! Don't substitute!

$1 + a + 2b + 3c + 2ab + 3ac + 6bc + 6abc$

$= 6abc + 2ab + 3ac + a + 6bc + 2b + 3c + 1$

$= 2ab(3c + 1) + a(3c + 1) + 2b(3c + 1) + 1(3c + 1)$

$= (2ab + a + 2b + 1)(3c + 1)$

$= [a(2b + 1) + 1(2b + 1)](3c + 1)$

$= (a + 1)(2b + 1)(3c + 1)$

$= (999 + 1)(2(666) + 1)(3(333) + 1)$

$= (1000)(1333)(1000)$

$=\huge \boxed{1333000000}$

• Replacing a = 3c and b = 2c , we find that: $1 + a + 2b + 3c + 2ab + 3ac + 6bc + 6abc = 36c^{3} + 33c^{2} + 10c + 1$ .
• Factoring this expression, we have: $36c^{3} + 33c^{2} + 10c + 1 = (3c + 1)^{2}(4c + 1)$ .
• Now, replacing c = 333 , we have:
  $(3\times333 + 1)^{2}(4\times333 + 1) =$ $(1000)^{2}(1333) =$ $13330000$ .
  

$(1+a+2b+3c+2ab+3ac+6bc+6abc) \\= 1+a+2b+2ab+3c+3ac+6bc+6abc \\= 1(1+a)+ 2b(1+a) + 3c (1+a) +6bc(1+a) \\= (1+a)(1+2b+3c+6bc) \\= (1+a) (1(1+2b)+3c(1+2b)) \\= (1+a)(1+3c) (1+2b) \\= 1000\times1000\times1333 \\= \boxed{1333000000}$

1 + a +2b + 3c + 2ab + 3ac + 6bc + 6abc

Factor 2b (1 + a) + 3c (1 + a) + (6bc + 1) (1 + a)

(1 + a) (2b + 3c + 6bc + 1)

(1 + a) [2b (3c + 1) + (3c + 1)]

(1 + a) (2b +1) (3c +1)

Then substitute the value of a, b, and c

(1 + 999) [2 (666) +1] [3(333) + 1] = 1 333 000 000

a = 3c, b = 2c

(1 + a) + (2b + 2ab) + (3c + 3ac) + (6bc + 6abc) =

(a + 1)(1 + 2b + 3c + 6bc) =

(a + 1)(1 +4c + 3c + 12c^2) = (a + 1)(1 +7c + 12c^2) = (a + 1)(1 + 3c)(1 + 4c) =

(a + 1) (a + 1)(1333) = (1000) (1000)(1333) = 1333000000

$1+a+2b+3c+2ab+3ac+6bc+6abc$

$(1+a)+2b(1+a)+3c(1+a)+6bc(1+a)$

$(1+a)(1+2b+3c+6bc)$

$Using, a=3c,b=2c , c = 333$

$(1+a)(1+2b+a+2ab)$

$(1+a)(1+a)(1+2b)$

$(1+a)(1+a)(1+a+c)$ = $\boxed{1000*1000*1333=1333000000}$

a = 3 c, b = 2 c

By substitution,

1 + 3c + 2 * 2c + 3c + 2 * 3c * 2c + 3 * 3c * c + 6 * 2c * c + 6 * 3c * 2c * c

which brings us to

1 + 3c + 4c + 3c + 12c * c + 9c * c + 12c * c + 36c * c * c

and then down to

1 + 10c + 33c * c + 36c * c * c

which I then plugged into a calculator. Not as glorious as factoring, but it got the answer fast enough.

1+a +2b+2ab+3c+3ac+6bc +6abc (1+a)+2b(1+a)+3c(1+a) +6bc(1+a) a=3c and b=2c (1+a)+4c(1+a)+3c(1+a)+12c^2(1+a) (1+a) (12c^2+4c+3c+1) 1000(4c(3c+1)+1(3c+1) ) =1000( (4c+1) (3c+1) ) 1000(1333) (1000) =1333000000

$1+a+2b+3c+2ab+3ac+6bc+6abc$

Rearrange as follows:

$=1+a+2b+2ab+3c+3ab+6bc+6abc$

Take common factors among the terms:

$=(1+a)+2b(1+a)+3c(1+a)+6bc(1+a)$

Factorise:

$=(1+a)(1+2b+3c+6bc)$

$=(1+a)[1(1+2b)+3c(1+2b)]$

Group all the factors:

$=(1+a)(1+2b)(1+3c)$

Substitute values:

$=(1+999)(1+2\cdot666)(1+3\cdot333)$

Simplify:

$=(1000)(1+1332)(1+999)$

$=(1000)(1333)(1000)$

$=\huge\boxed{1333000000}$

this is easy

1+a+2b+3c+2ab+3ac+6bc+6abc =1+a+2b+2ab+3c+3ac+6bc+6abc =(1+a)+2b(1+a)+3c(1+a)+6bc(1+a) =(1+a)(1+2b+3c+6bc) =(1+a){(1+2b)+3c(1+2b)} =(1+a)(1+2b)(1+3c) =(1000)(1333)(1000) =1333000000

(1+a)+2b(1+a)+3c(1+a)+6bc(1+a)

=(1+a)[1+2b+3c+6bc]

=1000[(1+2b)+3c(1+2b)]

=1000[(1+2b)(1+3c)]

=(1000)(1333)(1000)

=1333000000

$\begin{aligned} &\phantom{=}1+a+2b+3c+2ab+3ac+6bc+6abc\\ &=(1+a)+2b(1+a)+3c(1+a)+6bc(1+a)\\ &=(1+a)(1+2b+3c+6bc)\\ &=(1+a)((1+2b)+3c(1+2b))\\ &=(1+a)(1+2b)(1+3c)\\ &=(1+999)(1+2(666))(1+3(333))\\ &=(1000)(1333)(1000)\\ &=\boxed{1333000000} \end{aligned}$

We can rearrange the terms in the expression to make it easier to factorise. $\begin{aligned} 1 + a + 2b + 3c + 2ab + 3ac + 6bc + 6abc &= 6abc + 2ab + 3ac + a + 6bc + 2b + 3c + 1 \\ &= 2ab(3c + 1) + a(3c + 1) + 2b(3c + 1) + 3c + 1 \\ &= (2ab + a + 2b + 1)(3c + 1) \\ &= \left(a(2b + 1) + 2b + 1)(3c + 1\right) \\ &= (a + 1)(2b + 1)(3c + 1) \end{aligned}$ Then, substituting the values $\textcolor{#D61F06}{a = 999}$ , $\textcolor{#3D99F6}{b = 666}$ and $\textcolor{#20A900}{c = 333}$ gives: $\begin{aligned} (a + 1)(2b + 1)(3c + 1) &= (999 + 1)\left(2(666) + 1\right)\left(3(333) + 1\right) \\ &= (1000)(1333)(1000) \\ &= \boxed{1333000000} \end{aligned}$ Thus, the answer is 1333000000 .

$1+a+2b+3c+2ab+3ac+6bc+6abc$

Group all parts that are multiplied by $c$

$1+a+2b+2ab+(3c+3ac+6bc+6abc)$

Factor out $3c$

$1+a+2b+2ab+3c(1+a+2b+2ab)$

Now simplify $1+a+2b+2ab$ :

$(a+1)(2b+1)+3c(a+1)(2b+1)$

Factor out $(a+1)(2b+1)$

$(a+1)(2b+1)(3c + 1)$

Now just introduce the values for a, b and c then solve:

$(999+1)(2(666)+1)(3(333) + 1)$

$(1000)(1333)(1000)$

$1333000000$

The question is easy, since 1+a+2b+3c+2ab+3ac+6bc+6abc =(a+1)(2b+1)(3c+1) =(999+1)(2 666+1)(3 333+1) =1000x1333x1000 =1333000000

First, group the terms by their coefficients.

$(a+1)+(2b+2ab)+(3c+3ac)+(6bc+6abc)$

Next, factor a GCD from each set of parentheses.

$(a+1)+2b(a+1)+3c(1+a)+6bc(a+1)$

Since $a+1$ is common in each of the terms, we can factor it out, which leaves:

$(a+1)(1+2b+3c+6bc)$

Looking back at the values for a, b, and c, we can write $b$ as $2c$ and $a$ as $3c$ .

$(3c+1)(1+4c+3c+12c^2)$

$=(3c+1)(12c^2 + 7c +1)$

$=(3c+1)(4c+1)(3c+1)$

$=(3c+1)^{2}(4c+1)$

Finally, we plug in 333 for c, and we get $3c+1=1,000,000$ and $4c+1=1333$ . Multiplying 1,000,000 and 1333 together, we get $\boxed{1,333,000,000}$ .

=(1+a)+2b(1+a)+3c(1+a)+6bc(1+a) =(1+a)(1+2b+3c+6bc) =(1+a){1+2b+3c(1+2b)} =(1+a)(1+2b)(1+3c) =1000(1333)(1000) =1333000000

The answer is 1,333,000,000 - and the funny thing is, somewhere in my calculations I made an error, and I got 1,332,999,000

it easy but due to calculation errors one can get it right for third time.

a=3c and b=2c

then 1+a+2b+3c+2ab+3ac+6bc+6abc will be equal to

1+3c+4c+3c+12c c+9c c+12c c+36c c*c

=1+7c+33c^2+36c^3

=1+7(333)+33(333)(333)+36(333)(333)(333)=1333000000

1+a+2b+3c+2ab+3ac+6bc+6abc =1+a+2b+2ab+3c+3ac+6bc+6abc =(1+a)+2b(1+a)+3c(1+a)+6bc(1+a) =(1+a)(1+2b+3c+6bc) =(1+a){(1+2b)+3c(1+2b)} =(1+a)(1+2b)(1+3c) =(1000)(1333)(1000) =1333000000