Help me out!

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

I have tried to take (y+zx)3(y+z-x)^3 to the Left hand side and tried to use a3b3=(ab)(a2+b2+ab)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?

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

I am stuck in an another question, that is:

If x,yx,y and zz are real and unequal numbers, prove that: 2015x2+2015y2+6z2>2(2012xy+3yz+3xz)2015x^2+2015y^2+6z^2>2(2012xy+3yz+3xz)

Note by Aaryan Maheshwari
1 year, 11 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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 </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

To prove that it's an algebraic identity for k=24k=24, let X=y+zx,Y=z+xy,Z=x+yzX = y+z-x, Y = z+x-y, Z = x+y-z, and use the algebraic identity X3+Y3+Z3=3XYZ+(X+Y+Z)(X2+Y2+Z2XYYZXZ)X^3 + Y^3 + Z^3 = 3XYZ + (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-XZ) .

Pi Han Goh - 1 year, 11 months ago

Log in to reply

Thanks for your kind reply!

Aaryan Maheshwari - 1 year, 11 months ago

Log in to reply

Find the coefficient of each term

x3:1=1+1+1 x^3: 1 = -1 + 1 + 1
x2y:3=33+3 x^2 y: 3 = 3 -3 +3
xyz:6=666+k xyz: 6 = -6 - 6 - 6 + k

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

Calvin Lin Staff - 1 year, 11 months ago

Log in to reply

How can we find the coefficient?

Aaryan Maheshwari - 1 year, 11 months ago

Log in to reply

A bonus question:

If a=2015,b=2014 and c=12014,prove that\text{If}\space a=2015, b=2014\space \text{and}\space c=\frac{1}{2014}, \text{prove that} (a+b+c)3(a+bc)3(b+ca)3(c+ab)323abc=2015(a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3-23abc=2015

Aaryan Maheshwari - 1 year, 11 months ago

Log in to reply

Can I ask how to add hashtags?

Aaryan Maheshwari - 1 year, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...