# Basics Faster Than Skills

Factorize $$(b-c)^{6}+(c-a)^{6}+(a-b)^{6}-9(a-b)^{2}(b-c)^{2}(c-a)^{2}-2(a-b)^{3}(a-c)^{3}-2(b-c)^{3}(b-a)^{3}-2(c-a)^{3}(c-b)^{3}$$

• When you got the answer you must be surprised, feeling cheated.
• Think about the title carefully, in that there are two ways to solve it.

Note by Jason Snow
3 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Yes indeed...it looks quite symmetric doesn't it?

Let $$a-b=x, b-c=y, c-a=z$$

Then the expression becomes:

$$x^6+y^6+z^6-9x^2y^2z^2+2x^3y^3+2y^3z^3+2z^3x^3$$

$$= (x^3+y^3+z^3)^2-(3xyz)^2$$

$$= (x^3+y^3+z^3-3xyz)(x^3+y^3+z^3+3xyz)$$

Now we note that $$x+y+z=0$$ which means that $$x^3+y^3+z^3=3xyz$$ and our whole expression becomes 0

- 3 years, 1 month ago

That's why I said you'll be shocked :D, your method is faster than those I thought because I, in fact, want you guys to get it simplified step by step first, obviously you didn't jump into my trap :P

- 3 years, 1 month ago

0,same way as rohit.

- 3 years, 1 month ago

Aha, another brilliant person

- 3 years, 1 month ago

0

- 3 years, 1 month ago

Yeah, did it take you a long time? :D

- 3 years, 1 month ago

Certainly not....

- 3 years, 1 month ago

My advice to all regarding such long questions is to first try your luck with 0,and if it doesn't work,move to the next one

- 3 years, 1 month ago

Ha-ha, next time I'll try it, thanks for your nice method :D

- 3 years, 1 month ago