What is the product of 3 consecutive integers if the largest integer is (x+1)?

Question

Its hard.

Answer 1

Integers are x - 1 , x , x + 1
Product = (x - 1) (x) (x + 1)
Product = (x) (x - 1) (x + 1)
Product = (x) (x² - 1)
Product = x³ - x

Answer 2

If x + 1 is the largest integer, x and x - 1 are the other two integers (remember, they are consecutive).

So, (x - 1) . x . (x + 1) = x (x^2 - 1) = x^3 - x

Answer 3

Suppose the largest of 3 integers is equal to (x + 1). Then the previous two integers would be x and (x - 1). Therefore, multiplying them together,

x(x - 1)(x + 1)

Expand this. Since we have a product of conjugates, we have a difference of squares.

x(x^2 - 1)

Expanding,

x^3 - x

Answer 4

If largest integer is (x+1), then other integers would be x & (x-1). Their product would be
(x-1)*x*(x+1)
=x*(x-1)*(x+1)
since (x+a)(x-a) = x^2-a^2
=x*(x^2-1)
=x^3-x

Answer 5

the largest integer is

x + 1

the next one down consecutively is (x+1)-1 = x

and the remaining integer is (x+1)-2 = x -1

then to multiply them together

(x+1)x(x-1) = x (x^2 - 1)
= x^3 - x

then plug in whatever value for x that you consider right

Answer 6

Since x+1 is the largest the next one to it is x and the one after that is x-1.

(x+1) * x * (x-1) = x^3 - x

Answer 7

(x-1) x (x+1)
= x^3 -x

Answer 8

if the largest integer is (x+1)
then the one before is x
and the one before that is x-1
so the product is x(x-1)(x+1)
= x(x^2-1)
=x^3 - x

Answer 9

ha, ha, ha,

you have got to be kidding! why don't you ask why the theory of relativity is still in place after its been determined that it is wrong.

hey, here is a better question: how can you generate an unlimited supply of free energy utilizing excretion?

here is another goodie:

what is the result of multiplying the sum of the square of pi divided by its reciprocals square times the speed of light less the square root of gravity multiplied by the inverse of rho; where density is, (1000kg), gravity is (9.8), flow is (10 cu. m/sec.) and the head is 30.08 meters.