symmetric polynomials

fatorize the following polynomial f(x,y,z)=((x+y+z)^5)-(x^5+y^5+z^5).

#Algebra

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`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}`	$\boxed{123}$

Comments

5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx). Assume, x+y=0. Then x=-y and the expression automatically becomes 0. Thus by symmetry, y+z,z+x are also factors. Thus remaining factor is a quadratic factor which is symmetric in x,y,z. Assume it k1(x^2+y^2+z^2)+k2(xy+yz+zx), where k1 and k2 are the coefficients of the identity. Put random values of x,y,z and you will find out k1=k2=5.

Kushal Dey - 1 year, 2 months ago

Thank you soo much.

Gnanananda Shreyas - 1 year, 2 months ago

i got 3 factors i.e a-b,b-c,c-a but i camt find the last symmetric factors.

Gnanananda Shreyas - 1 year, 2 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}`	$\boxed{123}$