If x+y+z=0, show that x^{3}+y^{3}+z^{3}=3 x y z .

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Hindi

2. Polynomials

hard

If $x+y+z=0,$ show that $x^{3}+y^{3}+z^{3}=3 x y z$.

Option A

Option B

Option C

Option D

Since $x+y+z=0 $ $\therefore x+y=-z$

or $(x+y)^{3}=(-z)^{3}$ or $x^{3}+y^{3}+3 x y(x+y)=-z^{3}$

or $x^{3}+y^{3}+3 x y(-z)=-z^{3}$ $[\because x+y=(-z)]$

or $x^{3}+y^{3}-3 x y z=-z^{3}$ or $\left(x^{3}+y^{3}+z^{3}\right)-3 x y z=0$

or $\left(x^{3}+y^{3}+z^{3}\right)=3 x y z$

Hence, if $x+y+z=0,$ then $\left(x^{3}+y^{3}+z^{3}\right)=3 x y z$

Standard 9

Mathematics

hard

medium

easy

medium

medium