Do you know your abcs?

Calculus Level 3

E = lim ( a , b , c ) ( 1 , 1 , 1 ) a 3 + b 3 + c 3 3 a b c a 2 + b 2 + c 2 a b a c b c \mathscr{E}=\displaystyle \lim_{(a,b,c)\to(1,1,1) }\dfrac{a^3 + b^3 + c^3 -3abc}{a^2 +b^2 +c^2 - ab- ac -bc}

Find E \mathscr{E} .


The answer is 3.00.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Otto Bretscher
May 29, 2016

Using long division, we see that the expression simplifies to a + b + c a+b+c , so that the limit is 3 \boxed{3}

That's cheating!!! :) I intended for factoring to happen.

Hobart Pao - 5 years ago

Log in to reply

Long division = factoring. a 3 + b 3 + c 3 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 a b a c b c ) a^3 + b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc) is a common algebraic identity, which can be proven using Newton's identities .

Pi Han Goh - 5 years ago

Log in to reply

In the US, that's not a common identity. The way I approached a 3 + b 3 + c 3 3 a b c a^3 + b^3 + c^3 - 3abc was adding and subtracting 3 a 2 b + 3 a b 2 3a^2 b + 3ab^2 , then factoring. But it's probably common for Brilliant members anyway.

Hobart Pao - 5 years ago

Log in to reply

@Hobart Pao We practiced long division a lot ("ad nauseam") in High School back in Switzerland, decades ago. I have not used the technique much before I started to play on Brilliant... now it often comes in handy ;)

I have watched Americans (trying to) do long division, but, in my humble opinion, they don't do it right: They draw too many weird lines ;)

Otto Bretscher - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...