The functional equation identity, (assuming also $\,f(-x)=-f(x)\,$ for all $\,x$), $$ f(a)f(b)f(a\!-\!b) + f(b)f(c)f(b\!-\!c) + f(c)f(a)f(c\!-\!a) + f(a\!-\!b)f(b\!-\!c)f(c\!-\!a) = 0 \tag{1}$$ for all $\,a,b,c\,$ has solutions $f(x)=k_1\sin(k_2\,x)$ and $f(x)=k_1\tan(k_2\,x)\,$ with $\,k_1,k_...

When we break down the identity we have polynomial: $a^4 + b^4 + c^4 -2(a^2b^2 +a^2c^2 + b^2c^2)$ that we have to find zeros for. Obviously factor must be $a+b+c$ for this equation to be provable. But how to find factors,except dividing polynomial with $a+b+c$?

$$u_0(t)=0, \ u_1(t) = 1, \ u_n(t) = tu_{n-1}(t) - u_{n-2}(t)$$ I need to prove that $u_n(t)^2 - u_k(t)^2 = u_{n-k}(t)u_{n+k}(t), \ k =0,1,\dots,n$. I tried to prove statement by induction: Let $u_{k'}(t)^2 - u_k(t)^2 = u_{k'-k}(t)u_{k'+k}(t)$. Then: $$\begin{split}u_{k' + 1}(t)^2 - u_k(t)^2 &= (...

If $a^2,b^2,c^2$ are in Arithmetic Progression, prove that $b+c,c+a,a+b$ are in Harmonic Progression. Arithmetic mean: $b^2=\dfrac{a^2+c^2}{2}$ Harmonic mean: $c+a=\dfrac{2(b+c)(a+b)}{(b+c)+(a+b)}$ $\Rightarrow c+a=\dfrac{2ab+2b^2+2ac+2bc}{a+2b+c}$ Plug in $b^2=\dfrac{a^2+c^2}{2}$ $\Rightarrow c+...

If $a,b,c$ are in an Arithmetical Progression, and $b,c,d$ are in an Harmonical Progression, show that $a,\dfrac{c^2}{d},c$ are in H.P. and $b,\dfrac{ad}{b},d$ are also in H.P. Arithmetical mean: $a+c=2b$ $c=2b-a \tag{1}$ Harmonical mean: $\dfrac{2bd}{b+d}=c$ Thus $\dfrac{2bd}{b+d}=c \Rightarrow...

The question is in regards to the two lemniscatic elliptic functions, often called the 'sine lemniscate' and 'cosine lemniscate' functions. I have been trying to prove the following identity: \begin{align} \frac{sl(x) sl'(y) + sl'(x) sl(y)}{1 + sl(x)^2 sl(y)^2} \end{align} as equal to: \begin{ali...

Prove: if $ax^2-bx+c$ and $dx^3-bx+c$ have a common factor, then $a^3-abd+cd^2=0$. Not really sure how to proceed. I know that first I have to find the common factor. I guess it is $x-c$ because both expressions end with the constant $c$. Then plug in $x=c$ to get $ac^2-bc+c=0, c^3d-bc+c=0$. Henc...

Reading Wikipedia article on Diophantus, it says in a book that survived that he makes reference to a lost book called Porisms and the theorem stated in the title: the difference between the cubes of any 2 rationals can be expressed as the sum of the cubes of 2 rationals. Anyone point me to this ...

Let $$a_n=\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots$$ $$b_n=\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\cdots$$ $$c_n=\binom{n}{2}+\binom{n}{5}+\binom{n}{8}+\cdots$$ Then find the value of $${a_n^2}+{b_n^2}+{c_n^2}-a_nb_n-b_nc_n-c_na_n$$ I wrote all expressions in a summation form but couldn't ...

I want to write $P(x,y,z)=yx^{3}+zx^{3}+xy^{3}+zy^{3}+xz^{3}+yz^{3}$ in terms of elementary symmetric polynomials, but I'm getting stuck at the first step. I know I should follow the proof of the fundamental theorem of symmetric polynomials using the Newton identities. First I pick out the 'bi...

Problem The following identity is obvious: $$ \frac{1}{(x-y)(y-z)} + \frac{1}{(y-z)(z-x)} + \frac{1}{(z-x)(x-y)} = 0 $$ This post is for a trigonometric version in terms of cotangent: $$ \cot (x- y) \cot( y- z) + \cot( y-z) \cot( z - x) + \cot ( z- x) \cot ( x-y) = 1 $$ The following Mathematica ...

Here is a problem from Gelfand's Trigonometry: Let $\alpha, \beta, \gamma$ be any angle, show that $$\sin(\alpha -\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha-\gamma}{2}\right)\cos\left(\frac{\beta-\gamma}{2}\right).$$ I have t...

I have already made the observation that since the domain is real numbers, $a^2+b^2+c^2$ the three terms is always positive but $a^3+b^3+c^3$ since its to an odd power can have negative terms implying that at least one of $a$, $b$ or $c$ must be negative.

Resolved: The tag has been removed from all questions. What on Earth is the tag algebraic-identities for? It seems to me that questions containing at least one equal sign get randomly tagged with it, including questions that have nothing to do with algebra. The tag excerpt (there's no tag wi...