Question

If x3 + y3 + z3 = 6xyz and xyz = x + y + z, then the value of (x – y)2 + (y – z)2 + (z – x)2 is equal to​

Answer 1

Given:

$x^3 + y^3 + z^3 = 6xyz$ and

$xyz = x + y + z$

To find:

$(x - y)^2 + (y - z)^2 + (z - x)^2 = ?$

Solution:

First of all, let us expand the expression which is to be solved:

$(x - y)^2 + (y - z)^2 + (z - x)^2 \\\Rightarrow x^{2} +y^{2} -2xy+y^{2} +z^{2} -2yz+z^{2} +x^{2} -2zx \ \ \ (\because (a-b)^2 = a^2+b^2-2ab)\\\Rightarrow 2(x^{2} +y^{2}+z^{2} ) -2xy-2yz-2xz\\\Rightarrow 2(x^{2} +y^{2}+z^{2} -xy-yz-xz)$

So, we have to find the value of $2(x^{2} +y^{2}+z^{2} -xy-yz-xz)$

We know the Formula:

$a^3 + b^3 + c^3 - 3abc = (a + b + c )(a^2 + b^2 + c^2 -ab - ac -bc)$

Here, $\bold{a = x, b = y}\ and\ \bold{c = z}$

$\Rightarrow x^3 + y^3 + z^3 - 3xyz = (x + y + z )(x^2 + y^2 + z^2 -xy - yz -zx)$

Putting the given values:

$x^3 + y^3 + z^3 = 6xyz$ and $x + y + z= xyz$, the above expression becomes:

$\Rightarrow 6xyz- 3xyz = xyz(x^2 + y^2 + z^2 -xy - yz -zx)\\\Rightarrow 3xyz = xyz(x^2 + y^2 + z^2 -xy - yz -zx)\\\Rightarrow (x^2 + y^2 + z^2 -xy - yz -zx) = 3$

Multiplying the term with 2:

$2(x^2 + y^2 + z^2 -xy - yz -zx) = 3 \times 2\\\Rightarrow 2(x^2 + y^2 + z^2 -xy - yz -zx) = 6$

So, the answer is 6.