Question

If
a+b+c=4
and
a^2+b^2+c^2=10
and then
a^3+b^3+c^3=22
Then what does
a^4+b^4+c^4=?

Answer 1

I think what we have to do is use this
a+b+c=4
and
a^2+b^2+c^2=10
and then
a^3+b^3+c^3=22

To find what variable \[a,\ b,\ and\ c\ are\]

Am I correct @Imagine?

Answer 2

u r correct @supie

Answer 3

The and is 58 bcoz the distance between the first and to the second is 6 the second 17 then d last will be 28 making it 58

Answer 4

yes

Answer 5

@supie

Answer 6

Do you know the answer or is this a challenge?

Answer 7

I mean do you need the answer. or is it a challenge

Answer 8

I don't understand, how to add all them o get the answer.

Answer 9

Oh ok.

Answer 10

@jhonyy9

Answer 11

.

Answer 12

.-.

Answer 13

can u please close this because im tying to do hw and this is drawing my attention away from my work

Answer 14

this is my question? and you want me to close it?

Answer 15

yes please

Answer 16

I need this answered as well tho

Answer 17

they already helped right?

Answer 18

@Mercury?

Answer 19

think what we have to do is use this
a+b+c=4
and
a^2+b^2+c^2=10
and then
a^3+b^3+c^3=22

To find what variable
a, b, and c are

Answer 20

Thats what I said.

Answer 21

thats wat supie said

Answer 22

Still isn't solved tho

Answer 23

Smh

Answer 24

we cant tell the anwsers just help

Answer 25

smh

Answer 26

They can give you an answer if they back it up on why it's correct.

Answer 27

-_-

Answer 28

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Imagine
They can give you an answer if they back it up on why it's correct.
\(\color{#0cbb34}{\text{End of Quote}}\)
you deserve a badge for that

Answer 29

Oop- Thanks.

Answer 30

Bro I stumped Jhonny

Answer 31

Smh

Answer 32

a+b+c=4
and
a^2+b^2+c^2=10
and then
a^3+b^3+c^3=22
Then what does
a^4+b^4+c^4=?

a+b+c=4
a = 4-b-c
(4-b-c)^2+b^2+c^2=10
\[b^2+2bc+c^2-8b-8c+16+b^2+c^2=10\]
\[2b^2+2bc+2c^2-8b-8c+6=0\]
\[2(b^2+bc+c^2-4b-4c+3)=0\]


We also know that a^3+b^3+c^3=22
and a = 4-b-c
so we rewrite this
\[(4-b-c)^3+b^3+c^3=22\] simplify:
\[(-b^3-3b^2c-3bc^2-c^3+12b^2+24bc+12c^2-48b-48c+64)+ b^3 + c^2 = 22\]
\[(-3b^2c-3bc^2+12b^2+24bc+12c^2-48b-48c+64) = 22\]

Answer 33

NO WAY DID YOU JUST SOLVE THIS

Answer 34

So...
\[2(b^2+bc+c^2-4b-4c+3)=0\]
\[(-3b^2c-3bc^2+12b^2+24bc+12c^2-48b-48c+64) = 22\]
2 equations and 2 variables, from here you can solve for either b or c, and then plug that number back in, after that once you get b and c plug back into the first equation and solve for a

Answer 35

brooo

Answer 36

Not even jhonny solved this

Answer 37

Brooo

Answer 38

Once you get a, b, c then plug in values into
a^4+b^4+c^4=?

Answer 39

Except one problem, how in the world do we find b and c? I made the equations but like...

Answer 40

what is this problem? no way i am doing geometry rn

Answer 41

Nope.

Answer 42

This is Advanced Calc.

Answer 43

._.

Answer 44

hmm, I think I got those equations correct but i don't know how to solve for b and c because of all the variables having different squares and such, it might be doable but it will def require a lot of math. These are just my thoughts on this but I can't solve the whole thing without committing hours to this.

Answer 45

That's good enough for me, I'll fiqure it out from here. Thanks

Answer 46

<3

Answer 47

np