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Let $0 \le a, b, c \le 5$ be integers. For how many ordered triples $(a,b,c)$ is $a^2b+b^2c+c^2a-ab^2-bc^2-ca^2 = 0$?

 

THANKS!!!

 Mar 25, 2021
 #1
avatar+114087 
+1

please include the address of the first question.

 Mar 25, 2021
 #3
avatar+20 
+1

https://web2.0calc.com/questions/help-pls-d

CoolKid  Mar 25, 2021
 #2
avatar+114087 
+2

 

\(a^2b+b^2c+c^2a-ab^2-bc^2-ca^2 = 0\\ a^2b+b^2c+c^2a=ab^2+bc^2+ca^2 \\ \text{factor out b}\\ b(a^2+bc)+ac^2=b(c^2+ab)+ca^2\\ \)

Now, no matter what integer b is,   a =b

Convince yourself of that.....

 

Factor out c instead of b and what do you find?

Factor out a instead of b and what do you find?

 

Now you have all of your answer sets.

 Mar 25, 2021
 #4
avatar+20 
+1

Ok so I factored out each one:

 

$a(ab+c^2) + b^2c = ab^2 + bc^2 + ca^2$

 

and

 

$c(b^2 + ac) + a^2b = ab^2 + bc^2 + ca^2$

 

But how does this help me? Also thanks for you help. Really appreciate it! smiley

CoolKid  Mar 25, 2021
 #5
avatar+114087 
+1

I'll look at one of yours

 

\(a(ab+c^2) + b^2c = ab^2 + bc^2 + ca^2\\ a(ab+c^2) + b^2c = a(b^2+ca) + bc^2 \\ \text{I will assume a can equal anything at all}\\ \text{This means that}\\ ab+c^2=ca+b^2 \qquad and \qquad b^2c=bc^2\)

 

this would always be true if b=c

 

given the restiction that they all must be between 0 and 5 inclusive,

 

If a=0 these are all possibilities

a=0, b=0,c=0

a=0, b=1,c=1

a=0, b=2,c=2

a=0, b=3,c=3

a=0, b=4,c=4

a=0, b=5,c=5

 

If a=1 these are all possibilities

a=1, b=0,c=0

a=1, b=1,c=1

a=1, b=2,c=2

a=1, b=3,c=3

a=1, b=4,c=4

a=1, b=5,c=5

 

etc  a can be 0,1,2,3,4, or 5  that is 6 possible numbers for a,  each of these have 6 possibilities for b=c

That makes   6*6=36 possibilities here

 

Now do the other 2 factorisations, what do you get?

 Mar 26, 2021
 #6
avatar+20 
+1

YES I GOT It RIGHT THANKS SO MUCH MELODY!!! smiley

CoolKid  Mar 27, 2021

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