All categories

2 years ago

Show that x³ + y³ + z³ - 3xyz is a multiple of x+y+z

Mathematics

1 answer:

White raven [17]2 years ago

7 0

x³ + y³ + z³ - 3xyz is a multiple of x+y+z.

<h3>What is a multiple?</h3>

It should be noted that a multiple simply means a number that van be divided by another a certain number if times without a remainder.

Put x + y + z = 0

= x³ + y³ + z³ - 3xyx

= 0(x² + y² + z² - xt - yz - zx)

= x³ + y³ + z³ - 3xyz = 0.

x³ + y³ + z³ = 3xyz

Learn more about multiples on:

brainly.com/question/251701

#SPJ1

You might be interested in

What is 2 2/5 + 6 2/5?

Alekssandra [29.7K]

$2\dfrac{2}{5}+6\dfrac{2}{5}=(2+6)+\left(\dfrac{2}{5}+\dfrac{2}{5}\right)=8+\dfrac{2+2}{5}=8+\dfrac{4}{5}=8\dfrac{4}{5}$

6 0

4 years ago

The total daily sales, S, in the deli section of a supermarket is the sum of the purchases made by customers on a given day. (a)

MArishka [77]

Answer:

Step-by-step explanation:

FYI

Hope it will help

4 0

3 years ago

Cual es la raíz cuadrada de 35

nadya68 [22]

La raíz cuadrada de 35 creo que es 5

8 0

3 years ago

The area of an artist square canvas can hold 113 square square inches of paint. What is the approximate length of one side of th

Pavel [41]

Answer:

The approximate length of one side of the canvas (approximate to the nearest hundredth inch)= 10.63

Step-by-step explanation:

There are two ways to solve this problem . One is the use of the calculator and the other is the use of the fraction.

Now the square has equal sides length = breadth .

So we can simply find the length on one side by taking square root with the help of the calculator .

So when you press the symbol of square root after entering the number 113 we get the answer 10.63

Square = Length * breadth

Square = Length * length

Square = L²

One side =√ L²= L so √113= 10.63

The approximate length of one side of the canvas (approximate to the nearest hundredth inch)= 10.63

3 0

4 years ago

You flip a coin 5 times. in how many ways could u obtain at least one tail

Westkost [7]

Let's first think about how many possible outcomes there are to a series of coin flips. One that will help us here is that coin flips are independent - the outcome of one flip has no effect on the outcome of the others. What this means is that there are two possible outcomes for each flip: heads or tails. For an example with fewer coins, let's say we were flipping 2 instead of five.

- Flip 1 can either be heads or tails
- Flip 2 can either be heads or tails

So our possible outcomes are HH, HT, TH, and TT. There are two possible second flips for each of the possible first flips, or 2 x 2 = 4 total combinations of flips. Notice that only one of those combinations has zero tails - the combination with all heads.

If we were to flip a coin 5 times, we'd have two possible fifth flips for each of the two possible fourth flips for each of the two possible third flips for... it gets pretty hairy to describe in words, but I've attached a diagram so you can see how quickly it grows out of control. There are 2 x 2 x 2 x 2 x 2 or $2^5=32$ possible combinations of heads and tails! But, in fact, we don't even need to sort through these 32 combinations to answer our question. Every combination will contain at least one tail, except one: the combination which contains all heads (HHHHH). Which means the rest of the 31 must contain at least one tail.

This fact will stay the same regardless of the number of coin flips you make: the number of ways that contain at least one tail will always be the total number of combinations minus one (the case where all of the flips are heads).

5 0

4 years ago