Question

If x² + y² + z² = xy + yz + zx.
then the value of
(3x⁴ + 7y⁴+ 5z⁴)/5x²y² + 7y²z² - 3z²x² is?​

Answer 1

Answer:

$\frac{3x^4 + 7y^4 + 5z^4}{5x^2y^2+7y^2z^2-3z^2x^2} = \frac{5}{3}$

Step-by-step explanation:

Given :

If $x^2 + y^2 + z^2 = xy+yz+zx$

then the value of : $\frac{3x^4 + 7y^4 + 5z^4}{5x^2y^2+7y^2z^2-3z^2x^2}$

Solution :

If $x^2 + y^2 + z^2 = xy+yz+zx$,

then,

⇒ $x^2 + y^2 + z^2 - xy - yz - zx = 0$

By multiplyig & dividing by 2 , both the sides ,.

We get,

⇒ $\frac{1}{2}(2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx) = 0$

⇒ $\frac{1}{2}[(x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)] = 0$

⇒  $\frac{1}{2} [(x-y)^2 + (y-z)^2 + (z - x)^2]=0$

⇒ $(x-y)^2 + (y-z)^2 + (z - x)^2 = 0$

If sum of squares are zero,. then, the numbers must be 0,.

⇒ $x - y = 0$ , $y - z = 0$ , $z - x = 0$

⇒ $x = y$ , $y=z$ , $z=x$

⇒ $x=y=z$

By converting every term to x,.

$\frac{3x^4 + 7y^4 + 5z^4}{5x^2y^2+7y^2z^2-3z^2x^2}$

⇒ $\frac{3x^4 + 7x^4 + 5x^4}{5x^2(x^2) + 7(x^2)(x^2) - 3(x^2)(x^2)}$

⇒ $\frac{3x^4 + 7x^4 + 5x^4}{5x^4 + 7x^4 - 3x^4}$

⇒ $\frac{15x^4}{9x^4} = \frac{5}{3}$