Help me out!

Here is a tremendously huge time consuming and awkward looking question: $(x+y+z)^3=(y+z-x)^3+(z+x-y)^3+(x+y-z)^3+kxyz$ $\text{Find}\space k.$

I have tried to take $(y+z-x)^3$ to the Left hand side and tried to use $a^3-b^3=(a-b)(a^2+b^2+ab)$ , but I just kept on revolving and revolving and ended up being more confused than before.

Can anybody help me out by showing me how to factorize the expression or show me a simpler way?

$\huge\color{#D61F06}\text{Just a 12 year old math loving kid.}$ $\mathbb{\huge{\color{#456461}{\text{\#ASIMPLERWAY}}}}$

I am stuck in an another question, that is:

If $x,y$ and $z$ are real and unequal numbers, prove that: $2015x^2+2015y^2+6z^2>2(2012xy+3yz+3xz)$

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

Substitute $x=y=z=1$ and you will get $k=24$ .

To prove that it's an algebraic identity for $k=24$ , let $X = y+z-x, Y = z+x-y, Z = x+y-z$ , and use the algebraic identity $X^3 + Y^3 + Z^3 = 3XYZ + (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-XZ)$ .

Pi Han Goh - 3 years, 8 months ago

Thanks for your kind reply!

Aaryan Maheshwari - 3 years, 8 months ago

Find the coefficient of each term

$x^3: 1 = -1 + 1 + 1$
$x^2 y: 3 = 3 -3 +3$
$xyz: 6 = -6 - 6 - 6 + k$

Hence, $k = 24$ (and we can also conclude that we have an identity).

Calvin Lin Staff - 3 years, 8 months ago

How can we find the coefficient?

Aaryan Maheshwari - 3 years, 8 months ago

A bonus question:

$\text{If}\space a=2015, b=2014\space \text{and}\space c=\frac{1}{2014}, \text{prove that}$ $(a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3-23abc=2015$

Aaryan Maheshwari - 3 years, 8 months ago

Can I ask how to add hashtags?

Aaryan Maheshwari - 3 years, 8 months ago

Math	Appears 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}`	$\boxed{123}$