Math, asked by mansichoubisa9466, 1 month ago

x3 + y3 + z3 – 3xyz = 1/2( x + y + z ) [( x – y )2 + ( y - z )2 + ( z - x )2 ]

Answers

Answered by ZaraAntisera
0

Answer:

\mathrm{solve\:for\:x,\:x^3+y^3+z^3-3xyz=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\quad :\quad }

\mathrm{True\ for \ all \ x }

Step-by-step explanation:

\mathrm{x^3+y^3+z^3-3xyz=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}

\mathrm{x^3+y^3+z^3-3xyz=y^3-3xyz+x^3+z^3}

\mathrm{Subtract\:}y^3+z^3\mathrm{\:from\:both\:sides}

\mathrm{x^3+y^3+z^3-3xyz-\left(y^3+z^3\right)=y^3-3xyz+x^3+z^3-\left(y^3+z^3\right)}

\mathrm{x^3-3xyz=-3xyz+x^3}

\mathrm{Subtract\:}-3xyz+x^3\mathrm{\:from\:both\:sides}

\mathrm{x^3-3xyz-\left(-3xyz+x^3\right)=-3xyz+x^3-\left(-3xyz+x^3\right)}

0=0

\mathrm{Both\:sides\:are\:equal}

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