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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
3 weeks, 6 days 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 \(X^3 + Y^3 + Z^3 = 3XYZ + (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-XZ) \).

Pi Han Goh - 3 weeks, 5 days ago

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Thanks for your kind reply!

Ashok Dargar - 3 weeks, 4 days ago

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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 weeks, 5 days ago

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How can we find the coefficient?

Ashok Dargar - 3 weeks, 4 days ago

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Can I ask how to add hashtags?

Ashok Dargar - 3 weeks, 4 days ago

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A bonus question:

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

Ashok Dargar - 3 weeks, 4 days ago

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