×

# NMTC Problem 5a

If none of $$a,b,c,x,y,z$$ is zero and $$\frac { { x }^{ 2 }(y+z) }{ { a }^{ 3 } } =\frac { { y }^{ 2 }(z+x) }{ { b }^{ 3 } } =\frac { { z }^{ 2 }(x+y) }{ { c }^{ 3 } } =\frac { xyz }{ abc } =1$$.

Prove that $${a}^3+{b}^3+{c}^3+abc=0$$.

This a part of my set NMTC 2nd Level (Junior) held in 2014.

Note by Siddharth G
3 years, 2 months 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:

$$x^{2}(y+z)=a^{3}$$ and $$xyz=abc$$. Dividing the first equation by the second, we get $$x (\frac {1}{y}+\frac {1}{z})=\frac {a^{2}}{bc}$$. We can do this two more times with different variables to get two more similar equations. Multiplying all of them,

$$xyz (\frac {1}{y}+\frac {1}{z})(\frac {1}{y}+\frac {1}{x})(\frac {1}{x}+\frac {1}{z})=\frac {(x+y)(y+z)(x+z)}{xyz}=1$$ hence

$$(x+y)(y+z)(x+z)=xyz$$. Expanding and subtracting both sides by $$xyz$$, then substituting $$xyz$$ with $$abc$$, $$x^{2}(y+z)$$ with $$a^{3}$$ and so on gives the desired result.

- 3 years, 2 months ago