# The question is below?

## If $x + y + z = 0$ then prove that ${\left({x}^{2} + x y + {y}^{2}\right)}^{3} + {\left({y}^{2} + y z + {z}^{2}\right)}^{3} + {\left({z}^{2} + z x + {x}^{2}\right)}^{3} = 3 \left({x}^{2} + x y + {y}^{2}\right) \left({y}^{2} + y z + {z}^{2}\right) \left({z}^{2} + z x + {x}^{2}\right)$

Jul 11, 2018

Let

${x}^{2} + x y + {y}^{2} = a$

${y}^{2} + y z + {z}^{2} = b$
and

${z}^{2} + z x + {x}^{2} = c$

So $a - b = {x}^{2} - {z}^{2} + x y - y z$

$= \left(x - z\right) \left(x + z\right) + y \left(x - z\right)$

$= \left(x - z\right) \left(x + z + y\right) = \left(x - z\right) \times 0 = 0$

Similarly we can get

$b - c = 0 \mathmr{and} c - a = 0$

Hence $a = b = c$

So$L H S = {\left({x}^{2} + x y + {y}^{2}\right)}^{3} + {\left({y}^{2} + y z + {z}^{2}\right)}^{3} + {\left({z}^{2} + z x + {x}^{2}\right)}^{3}$

$= {a}^{3} + {b}^{3} + {c}^{3} = {a}^{3} + {a}^{3} + {a}^{3} = 3 {a}^{3}$

And

$R H S = 3 a b c = 3 a \times a \times a = 3 {a}^{3}$

So $L H S = R H S$

Jul 12, 2018

Please refer to a Second Proof in Explanation.

#### Explanation:

Prerequisite : $a = b = c \Rightarrow {a}^{3} + {b}^{3} + {c}^{3} = 3 a b c \ldots \left(\star\right)$.

Let, a=x^2+xy+y^2, b=y^2+yz+z^2, &, c=z^2+zx+x^2.

Hence, $\left(a - b\right) = \underline{{x}^{2} - {z}^{2}} + \underline{x y - y z}$,

$= \underline{\left(x - z\right)} \left(x + z\right) + y \underline{\left(x - z\right)}$,

$= \left(x - z\right) \left(x + z + y\right)$,

=(x-z)(0).......................................[because," Given]".

$\Rightarrow a - b = 0 , \mathmr{and} , a = b$.

Similarly, we can show that, $b = c$.

Altogether, $a = b = c$.

$\therefore \text{ by } \left(\star\right) , {a}^{3} + {b}^{3} + {c}^{3} = 3 a b c , i . e . ,$

${\left({x}^{2} + x y + {y}^{2}\right)}^{3} + {\left({y}^{2} + y z + {z}^{2}\right)}^{3} + {\left({z}^{2} + z x + {x}^{2}\right)}^{3}$,

$= 3 \left({x}^{2} + x y + {y}^{2}\right) \left({y}^{2} + y z + {z}^{2}\right) \left({z}^{2} + z x + {x}^{2}\right)$.

$\textcolor{red}{\text{Enjoy Maths!}}$