x^{4}/(y^{2}z^{2}) + y^{4}/(z^{2}x^{2}) + z^{4}/x^{2}y^{2} = 3, then find the value of (x^{2} + y^{2} + z^{2})^{3}

Option 3 : 27

**Given :**

x^{4}/(y^{2}z^{2}) + y^{4}/(z^{2}x^{2}) + z^{4}/x^{2}y^{2} = 3

**Calculation:**

value putting,

Let x = 1, y = 1 and z = 1

Then

x4/(y2z2) + y4/(z2x2) + z4/x2y2 = 3

From left hand side

x4/(y2z2) + y4/(z2x2) + z4/x2y2

⇒ 1 + 1 + 1

⇒ 3 = R.H.S

(x2 + y2 + z2)^{3}

⇒ (1 + 1 + 1)^{3}

⇒ 3^{3} = 27

**∴ The value of (x2 + y2 + z2)3 is 27.**

__Mistake Points__

a3 + b3 + c3 = 3abc at (a + b + c) = 0

Calculations :

x4/(y2z2) + y4/(z2x2) + z4/x2y2 = 3

(x6 + y6 + z6)/(x2y2z2) = 3

x6 + y6 + z6 = 3x2y2z2

Comparing it with the equation a3 + b3 + c3 = 3abc , we get

x2 + y2 + z2 never equal to zero

Because summation of square term never gives zero value.