what's 40 x 40 x 40 please​

1: [tex]3n^{2}+9n+6[/tex]

notice that each part is divisible by 3

[tex]3n^{2}[/tex] ÷ 3 = [tex]n^{2}[/tex]

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes [tex]3(n^{2} +3n+2)[/tex]

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

[tex]3(n^{2} +2n+n+2)[/tex]

Now that it's rewritten, you can factor out n + 2 from the equation.

the answer is

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is [tex]n^{2} +2n+n+2[/tex] and then each of those by 3, which is [tex]3n^{2} +6n+3n+6[/tex] or [tex]3n^{2}+9n+6[/tex], our origional equation

2: [tex]28+x^{2} -11x[/tex]

So I rewrote this as [tex]x^{2} -11x+28[/tex] (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

[tex]x^{2} -4x-7x+28[/tex]

now we can factor out x from the first expression and -7 from the second

[tex]x(x-4)-7(x-4)[/tex]

and lastly you factor out x-4,

which would give you

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to [tex]x^{2} -11x+28[/tex]

3: [tex]9x^{2} -12x+4[/tex]

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as [tex]3^{2}[/tex] and 4 as [tex]2^{2}[/tex], the equation becomes

[tex]3^{2} x^{2} -12x+2^{2}[/tex]

now that [tex]3^{2} x^{2}[/tex] is ugly so it can be turned into [tex](3x)^{2}[/tex]

and -12x can be rewritten as [tex]-2*3x*2[/tex]

so our equation now looks like [tex](3x)^2-2*3x*2+2^{2}[/tex]

There's a rule that says [tex]a^{2} -2ab+b^{2} = (a-b)^{2}[/tex]

In our case, a=3x and b=2

so the final answer is

[tex](3x-2)^2[/tex]