# 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?

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

I am stuck in an another question, that is:

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

Note by Ashok Dargar
9 months, 3 weeks ago

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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 $$X^3 + Y^3 + Z^3 = 3XYZ + (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-XZ)$$.

- 9 months, 3 weeks ago

- 9 months, 3 weeks 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).

Staff - 9 months, 3 weeks ago

How can we find the coefficient?

- 9 months, 3 weeks ago

- 9 months, 3 weeks ago

A bonus question:

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

- 9 months, 3 weeks ago