Question

If
$x + y + z = 0$
then show that
${x}^{4} + {y}^{4} + {z}^{4} = 2( {x}^{2} {y}^{2} + {y}^{2} {z}^{2} + {z}^{2} {x}^{2} )$
​

Answer 1

Step-by-step explanation:

this is proved that left hand side = right hand side

Answer 2

Step-by-step explanation:

We have,

x + y + z = 0

Squaring both sides,

(x + y + z)² = 0²

Using the identity,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

(x + y + z)² = 0²

x² + y² + z² + 2(xy + yz + xz) = 0

x² + y² + z² = -2(xy + yz + xz)

Again Squaring both sides,

(x² + y² + z²)² = [-2(xy + yz + xz)]²

Using the identity,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

(x²)² + (y²)² + (z²)² + 2(x²y² + y²z² + x²z²)

= [(-2)²(xy + yz + xz)²]

x⁴ + y⁴ + z⁴ + 2(x²y² + y²z² + x²z²) = [4(xy + yz + xz)²]

x⁴ + y⁴ + z⁴ = [4(xy + yz + xz)²] - 2(x²y² + y²z² + x²z²)

Using the identity again,

x⁴ + y⁴ + z⁴ = [4(x²y² + y²z² + x²z² + 2(xy × yz + yz × xz + xy × xz)] - 2(x²y² + y²z² + x²z²)

x⁴ + y⁴ + z⁴ = [4(x²y² + y²z² + x²z² + 2(xy²z + xyz² + x²yz))] - 2(x²y² + y²z² + x²z²)

x⁴ + y⁴ + z⁴ = 4x²y² + 4y²z² + 4x²z² + 8(xy²z + xyz² + x²yz) - 2x²y² - 2y²z² - 2x²z²

x⁴ + y⁴ + z⁴ = 4x²y² + 4y²z² + 4x²z² - 2x²y² - 2y²z² - 2x²z² + 8(xy²z + xyz² + x²yz)

x⁴ + y⁴ + z⁴ = 2x²y² + 2y²z² + 2x²z² + 8(xy²z + xyz² + x²yz)

x⁴ + y⁴ + z⁴ = 2(x²y² + y²z² + x²z²) + 8xyz(y + z + x)

x⁴ + y⁴ + z⁴ = 2(x²y² + y²z² + x²z²) + 8xyz(x + y + z)

But we know that,

x + y + z = 0

So,

x⁴ + y⁴ + z⁴ = 2(x²y² + y²z² + x²z²) + 8xyz(0)

x⁴ + y⁴ + z⁴ = 2(x²y² + y²z² + x²z²) + 0

x⁴ + y⁴ + z⁴ = 2(x²y² + y²z² + x²z²)

Hence proved

Hope it helped and believing you understood it........All the best