+0

I'm going to repost since no one answered (except a troller)

0
135
6
+20

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
+114087
+1

Mar 25, 2021
#3
+20
+1

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

CoolKid  Mar 25, 2021
#2
+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?

Mar 25, 2021
#4
+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!

CoolKid  Mar 25, 2021
#5
+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
+20
+1

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

CoolKid  Mar 27, 2021